\frac{(i - 1)!}{(r - i + 1)(r - i)\cdots (r - 2i + 2)}.
\]

Putting these determinant computations together in Cramer's formula, we get

\begin{equation}\label{value for c in geq}
x_i = (-1)^{i - 1}\frac{(i - 1)!}{(r - i + 1)(r - i)\cdots (r - 2i + 2)} = (-1)^{i - 1} \frac{1}{i{r - i + 1 \choose i}}.
\end{equation}

Substituting this value of $x_i$ in equation (<ref>), we get

\[
0 \equiv (-1)^{i}i {r - i + 1 \choose i}\left[\sum_{a = 0}^{p - 1}p^{-1}\cL(z - a)^{r - i + 1}\mathbbm{1}_{a + p\Zp}\right] - z^{r - i + 1}\mathbbm{1}_{\Zp}.
\]

We show that this equation is the image of $-w(\lambda_i T_{-1, 0} - 1)\llbracket \id, X^{i - 1}Y^{r - i + 1}\rrbracket$ under the surjection $\IZind a^{i - 1}d^{r - i + 1} \twoheadrightarrow F_{2i - 2, \> 2i - 1}$.
Clearly, $w\cdot z^{i - 1}\mathbbm{1}_{p\Zp} = -z^{r - i + 1}\mathbbm{1}_{\Zp}$. Next, using
(<ref>) we have

\begin{eqnarray*}
-wT_{-1, 0}\llbracket \id, X^{i - 1}Y^{r - i + 1}\rrbracket & = & -\sum_{\lambda \in I_1}\left\llbracket \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}p & -\lambda \\ 0 & 1\end{pmatrix}, X^{i - 1}Y^{r - i + 1}\right\rrbracket \\
& \mapsto & -\sum_{\lambda \in I_1}\begin{pmatrix}0 & 1 \\ p & -\lambda\end{pmatrix}z^{i - 1}\mathbbm{1}_{p\Zp} \\
& = & -\sum_{\lambda \in I_1}p^{i - 1 - r/2}(z - \lambda)^{r - i + 1}\mathbbm{1}_{p\Zp}\left(\frac{p}{z - \lambda}\right) \\
& = & \sum_{\lambda \in I_1} p^{i - r/2 - 1}(z - \lambda)^{r - i + 1}\mathbbm{1}_{\lambda + p\Zp}.
\end{eqnarray*}

Therefore we have proved that the surjection $\IZind a^{i - 1}d^{r - i + 1} \twoheadrightarrow F_{2i - 2, \> 2i - 1}$ factors as

\[
\IZind a^{i - 1}d^{r - i + 1} \twoheadrightarrow \frac{\IZind a^{i - 1}d^{r - i + 1}}{\im(\lambda_iT_{-1, 0} - 1)} \twoheadrightarrow F_{2i - 2, \> 2i - 1}.
\]

Next assume that $\nu = i - r/2$. Consider the following exact sequences

\begin{eqnarray}
\label{rectangle}
\begin{tikzcd}
0 \arrow[r] & \im T_{-1, 0} \arrow[r] \arrow[d, two heads] & \IZind a^{i - 1}d^{r - i + 1} \arrow[r]\arrow[d, two heads] & \dfrac{\IZind a^{i - 1}d^{r - i + 1}}{\im T_{-1, 0}} \arrow[r]\arrow[d, two heads] & 0 \\
0 \arrow[r] & S \arrow[r] & F_{2i - 2, \> 2i - 1} \arrow[r] & Q \arrow[r] & 0,
\end{tikzcd}
\end{eqnarray}

where $S$ (for submodule) is the image of $\im T_{-1, 0}$ under the surjection $\IZind a^{i - 1}d^{r - i + 1} \twoheadrightarrow F_{2i - 2, \> 2i - 1}$ and $Q$ (for quotient) is $F_{2i - 2, \> 2i - 1}/S$. We know that the subspace $\im (\lambda_i T_{-1, 0} - 1)$ maps to $0$ under the middle vertical map in the display above. Since $\im (\lambda_i T_{-1, 0} - 1)$ maps surjectively onto $\dfrac{\IZind a^{i - 1}d^{r - i + 1}}{\im T_{-1, 0}}$ the right vertical map is $0$. Therefore we get a surjection $\im T_{-1, 0} \twoheadrightarrow F_{2i - 2, \> 2i - 1}$. First assume that $a^{i - 1}d^{r - i + 1}$ does not factor through $\omega$. For $1 \leq r \leq p - 1$ and $1 \leq i \leq \lceil r/2 \rceil - 1$, the character $a^{i - 1}d^{r - i + 1}$ factors through $\omega$ only when $i = 1$ and $r = p - 1$. Using the first isomorphism theorem and <cit.>, we get a surjection

\[
\frac{\IZind a^{i - 1}d^{r - i + 1}}{\im T_{1, 2}} \twoheadrightarrow F_{2i - 2, \> 2i - 1}
\]

obtained by pre-composing the first surjection above with $T_{-1,0}$. Since
$T_{-1,0} ( \im (\lambda_i T_{-1, 0} - 1) ) \subset   \im (\lambda_i T_{-1, 0} - 1)$
the last surjection factors through

\[
\frac{\IZind a^{i - 1}d^{r - i + 1}}{\im T_{1, 2} + \im (\lambda_i T_{-1, 0} - 1)} \twoheadrightarrow F_{2i - 2, \> 2i - 1}.
\]

Since $\lambda_i$ is a unit, the space on the left is the same

\[
\frac{\IZind a^{i - 1}d^{r - i + 1}}{\im T_{1, 2} + \im (T_{-1, 0} - \lambda_i^{-1})}
\simeq \pi([2i - 2 - r], \lambda_i^{-1}, \omega^{r - i + 1}),
\]

where $[2i - 2 - r] = 0, 1, \ldots, p - 2$ represents the class of $2i - 2 - r$ modulo $p - 1$, by
(<ref>), noting that
$[2i - 2 - r] \neq 0$ since $(i,r) \neq (1,p-1)$.

Now assume that $(i, r) = (1, p - 1)$. We remark that in this case $k = p+1$ and so this case was not considered in [9]. Even in our approach involving the
Iwahori mod $p$ LLC a new complication arises: we need to use the Hecke operator $T_{1,0}$ in the non-commutative Iwahori-Hecke algebra to analyze the subquotient $F_{0,1}$
Applying Proposition $\ref{nu geq}$ with $n = r = p - 1$, we get

\begin{eqnarray*}
0 & \equiv & \sum_{a = 0}^{p - 1}p^{-1}\cL(z - a)^r\mathbbm{1}_{a + p\Zp} - p^{-1}\cL \mathbbm{1}_{\Zp} + \sum_{a = 0}^{p - 1} \frac{a^r}{r}\mathbbm{1}_{a + p\Zp}.
\end{eqnarray*}

Here the third function on the right side of the congruence in the proposition simplifies to the
last term above by the remarks made in the first paragraph of the proof.
Multiplying this equation by $-r$ and using the definition of $\lambda_i$ in the statement of the

we get

\begin{eqnarray}\label{Crude final congruence for i,r = 1, p-1}
0 \equiv \lambda_i\left[\sum_{a = 0}^{p - 1}p^{-r/2}(z - a)^r\mathbbm{1}_{a + p\Zp} - p^{-r/2}\mathbbm{1}_{\Zp}\right] - \sum_{a = 0}^{p - 1}a^r\mathbbm{1}_{a + p\Zp}.
\end{eqnarray}

By the remarks at the beginning of the proof

\[
z^r\mathbbm{1}_{\Zp} = \sum_{a = 0}^{p - 1}\sum_{j = 0}^{r}{r \choose j}a^j(z - a)^{r - j}\mathbbm{1}_{a + p\Zp} \equiv \sum_{a = 0}^{p - 1}\left[a^r + (z - a)^r\right]\mathbbm{1}_{a + p\Zp} \equiv \sum_{a = 0}^{p - 1}a^r\mathbbm{1}_{a + p\Zp}.
\]

For the final congruence above, we have used Lemma <ref>. Substituting this in equation (<ref>), we get

\[
0 \equiv \lambda_i\left[\sum_{a = 0}^{p - 1}p^{-r/2}(z - a)^r\mathbbm{1}_{a + p\Zp} - p^{-r/2}\mathbbm{1}_{\Zp}\right] - z^r\mathbbm{1}_{\Zp}.
\]

Using (<ref>), we see exactly as before that the right side of the equation above is the image of the function $-w[\lambda_i(T_{-1, 0} + T_{1, 0}) - 1]\llbracket \id, Y^{r}\rrbracket$ under the surjection $\IZind d^{p - 1} \twoheadrightarrow F_{0, 1}$. Therefore the surjection $\IZind 1 \twoheadrightarrow F_{0, 1}$ factors as

\[
\IZind 1 \twoheadrightarrow \frac{\IZind 1}{\im(\lambda_i(T_{-1, 0} + T_{1, 0}) - 1)} \twoheadrightarrow F_{0, 1}.
\]

Now assume that $\nu = 1 - r/2$. Using Lemma <ref>, we see that this surjection further gives us

\[
\frac{\IZind 1}{\im(T_{1, 2}T_{1, 0}) + \im ((T_{-1, 0} + T_{1, 0}) - \lambda_{i}^{-1})} \twoheadrightarrow F_{0, 1}.
\]

By (<ref>), and noting $T_{1,0}$ is an automorphism, the space on the left is

\[
%                          \frac{\IZind 1}{\im(T_{1, 2}T_{1, 0}) + \im ((T_{-1, 0} + T_{1, 0}) - \lambda_i^{-1})} \simeq
\pi(0, \lambda_{i}^{-1}, 1) = \pi([2i - 2 - r], \lambda_i^{-1}, \omega^{r - i + 1}).
\]

Therefore we have proved the theorem.

§.§ The $\nu \leq i - r/2$ cases

In this section, we prove that if $\nu \leq i - r/2$ for $i = 1, \> 2, \> \ldots, \> \lceil r/2 \rceil - 1$, then the map $\IZind a^i d^{r - i} \twoheadrightarrow F_{2i, \> 2i + 1}$ factors as

\[
\IZind a^i d^{r - i} \twoheadrightarrow \frac{\IZind a^i d^{r - i}}{\im(\lambda_i^{-1} T_{1, \> 2} - 1)} \twoheadrightarrow F_{2i, \> 2i + 1},
\]

where $\lambda_i = (-1)^i i {r - i + 1 \choose i}p^{r/2 - i}\cL$. Moreover for $\nu = i - r/2$, the second map in the display above induces a surjection

\[
\pi(r - 2i, \lambda_i, \omega^i) \twoheadrightarrow F_{2i, \> 2i + 1}.
\]

Let $k/2 < n \leq k - 2 \leq p - 1$. For $\nu \leq 1 + r/2 - n$, set

\[
g(z) = p^x\left[\sum_{i \in I} \lambda_i (z - i)^n\logL(z - i)\right],
\]

where $x \in \bQ$ with $x + \nu = r/2 - n$, $I = \{0, 1, \ldots, n, p\}$ and the $\lambda_i \in \zp$
are as in Lemma <ref> (2).

Then, we have

\[
g(z) \equiv p^{1 + x}\sum_{a = 1}^{p - 1}a^{-1}z^n\mathbbm{1}_{a + p\Zp} + \sum_{j = \lceil r/2 \rceil}^{n - 1}(-1)^{n - j + 1}{n \choose j}p^{x + n - j}\cL z^j \mathbbm{1}_{p\Zp} \mod \pi \latticeL{k}.
\]

Write $g(z) = g(z)\mathbbm{1}_{\Zp} + g(z)\mathbbm{1}_{\Qp \setminus \Zp}$. By Lemma <ref> (2), we have

\begin{eqnarray}\label{Identities concerning the coefficients in leq}
\sum_{i \in I}\lambda_i i^{j} = 0, \text{ for } 0 \leq j \leq n,
\> \lambda_0 \equiv 1 \!\! \mod p, \> \lambda_i \equiv 0 \!\! \mod p \text{ for }
1 \leq i \leq n  \text{ and } \lambda_p = -1.
\end{eqnarray}

Since the summation identity in (<ref>) is also true
for $j = n$, the $\logL(z)$ term dies and so

\[
w\cdot g(z)\mathbbm{1}_{\Qp \setminus \Zp} = \sum_{i \in I}p^x\lambda_iz^{r - n}(1 - zi)^n\logL(1 - zi)\mathbbm{1}_{p\Zp}.
\]

Lemma <ref> implies that $w\cdot g(z) \mathbbm{1}_{\Qp \setminus \Zp} \equiv 0 \mod \pi \latticeL{k}$.

Next, using Lemma <ref> (1) we see that $g(z)\mathbbm{1}_{\Zp} \equiv g_2(z) \mod \pi \latticeL{k}$. We first claim that $g_2(z) - g_1(z) \equiv 0 \mod \pi \latticeL{k}$. Working as in the proof of Proposition <ref>, we see that the coefficient of $(z - a - \alpha p)^j \mathbbm{1}_{a + \alpha p + p^2 \Zp}$ in $g_2(z) - g_1(z)$ is

\begin{eqnarray}\label{jth summand in g2 - g1 in leq}
&& {n \choose j}
\sum_{i \in I}p^x \lambda_i  \Big[(a + \alpha p - i)^{n - j}\logL(a + \alpha p - i) - (a - i)^{n - j}\logL(a - i) \\
&& - {n - j \choose 1}(\alpha p)(a - i)^{n - j - 1}\logL(a - i) - \cdots - {n - j \choose n - j - 1}(\alpha p)^{n - j - 1}(a - i)\logL(a -  i)\Big]. \nonumber
\end{eqnarray}

By Lemma <ref>, to prove the claim it suffices to show
that this coefficient is congruent to $0$ modulo $p^{r/2 - j}\pi$.

First assume that $a \neq 0$. For $i \neq a$, the $i^{\mathrm{th}}$ summand in (<ref>)
can be written as

\[
% {n \choose j}
p^x\lambda_i\big[(a - i + \alpha p)^{n - j}\logL(1 + (a - i)^{-1}\alpha p) + (\alpha p)^{n - j}\logL(a - i)\big].
\]

The condition $\nu \leq 1 + r/2 - n$ and the equation $x + \nu = r/2 - n$ implies that $x \geq - 1$. Therefore the valuation of the last term in the expression above is greater than or equal to $n - j > r/2 - j$. So we drop the last term modulo $p^{r/2 - j}\pi$. Moreover, expanding $(a - i + \alpha p)^{n - j}$ and the logarithm, we get

\begin{eqnarray}\label{a neq 0 and zi neq a term}
% {n \choose j}
p^x\lambda_i\big[(a - i)^{n - j - 1}\alpha p + c_{n - j - 2}(a - i)^{n - j - 2}(\alpha p)^2 + \cdots + c_1(a - i)(\alpha p)^{n - j - 1}\big]
\!\!\!\! \mod p^{r/2 - j}\pi,
\end{eqnarray}

for some $c_1,\ldots, c_{n - j - 2} \in \Zp$.
Next assume that $i = a$. Then the $i^{\mathrm{th}}$ summand
in expression (<ref>) can be written as

\[
% {n \choose j}
p^x\lambda_i(\alpha p)^{n - j}\logL(\alpha p).
\]

The valuation of this term is greater than or equal to $1 + r/2 - j > r/2 - j$ since $p \mid \lambda_a$ by (<ref>). Therefore the $i=a$ summand
in expression (<ref>) is congruent to $0$ modulo $p^{r/2 - j}\pi$.

Using the summation identities in (<ref>)
and expression (<ref>), we see that the coefficient of $(z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2 \Zp}$ in $g_2(z) - g_1(z)$ is $0$ modulo $p^{r/2 - j}\pi$ if $a \neq 0$.

Now assume that $a = 0$. For $i \neq 0, \> p$, the $i^{\mathrm{th}}$ summand in expression (<ref>) is also given by (<ref>). Making a simplification similar to the one made in claim (<ref>), we see that the sum of the $i = 0$ and $i = p$ summands
in expression (<ref>) is

\begin{eqnarray*}
% {n \choose j}
p^x\left\{\left[(\alpha p)^{n - j}\cL\right] - \left[(\alpha p - p)^{n - j}\cL - (-p)^{n - j}\cL - {n - j \choose 1}(\alpha p)(-p)^{n - j - 1}\cL \right.\right.\\
\left.\left. - \cdots - {n - j \choose n - j - 1}(\alpha p)^{n - j - 1}(-p)\cL\right]\right\} \mod p^{r/2 - j}\pi.
\end{eqnarray*}

This expression is clearly $0$. Now using the identity concerning the sum in (<ref>), we see that the coefficient of $(z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2 \Zp}$ in $g_2(z) - g_1(z)$ is

\[
-{n \choose j}p^x\lambda_p\left[(-p)^{n - j - 1}\alpha p + c_{n - j - 2}(-p)^{n - j - 2}(\alpha p)^2 + \cdots + c_1(-p)(\alpha p)^{n - j - 1}\right] \mod p^{r/2 - j}\pi.
\]

Each term in the expression above has valuation greater than or equal to $n - j - 1 > r/2 - j$. Therefore we have proved that the coefficient of $(z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2\Zp}$ in $g_2(z) - g_1(z)$ is $0$ modulo $p^{r/2 - j}\pi$ if $a = 0$.

This shows that $g(z) \equiv g_1(z) \!\! \mod \pi \latticeL{k}$. Next, we simplify $g_1(z)$. Recall that

\begin{eqnarray}\label{g1 in leq}
g_1(z) = \sum_{a = 0}^{p - 1}\left[\sum_{j = 0}^{n - 1}\frac{g^{(j)}(a)}{j!}(z - a)^j\right]\mathbbm{1}_{a + p\Zp},
\end{eqnarray}

\begin{eqnarray}\label{jth summand in g1 in leq}
\frac{g^{j}(a)}{j!} = \sum_{i \in I}{n \choose j}p^x\lambda_i(a - i)^{n - j}\logL(a - i).
\end{eqnarray}

First assume that $a \neq 0$. The $i \neq 0, a, p$ summands on the right side of equation (<ref>) are congruent to $0$ modulo $\pi$ since $a \not \equiv i \!\! \mod p$ and $p \mid \lambda_i$ for $i \neq 0, p$. The $i = a$ term is equal to $0$. The sum of the $i = 0$ and $i = p$ terms
in (<ref>)

is congruent to

\[
{n \choose j}p^x\left[a^{n - j}\logL(a) - (a - p)^{n - j}\logL(a - p)\right] \mod \pi
\]

since, by (<ref>), $\lambda_0 = 1 \!\! \mod p$
and $\lambda_p = -1$.
Expanding $(a - p)^{n - j}$ and dropping the terms divisible by $p$ since $x \geq -1$ and $p \> \vert \> \logL(a - p)$, this equals

\[
{n \choose j}p^x(-a^{n - j})\logL(1 - a^{-1}p) \mod \pi.
\]

Expanding $\logL(1 - a^{-1}p)$ using the usual Taylor series and dropping the terms that are congruent to $0$ mod $\pi$, we see that the sum of the $i = 0$ and $p$ summands in (<ref>) is

\[
{n \choose j}p^{1 + x}a^{n - j - 1} \mod \pi.
\]

Therefore for $a \neq 0$, we get

\begin{eqnarray}\label{jth term in a neq 0 in g1 in leq}
\frac{g^{(j)}(a)}{j!} \equiv {n \choose j}p^{1 + x}a^{n - j - 1} \mod \pi.
\end{eqnarray}

Next, assume that $a = 0$. Then again the $i \neq 0, \> p$ summands in
(<ref>) are congruent to $0$ modulo $\pi$ by the same reasoning as in the $a \neq 0$ case above. The $i = 0$ summand is $0$. The $i = p$ summand is

\[
-{n \choose j}p^x(-p)^{n - j}\cL.
\]

Therefore for $a = 0$, we have

\begin{eqnarray}\label{jth term in a = 0 in g1 in leq}
\frac{g^{(j)}(0)}{j!} \equiv (-1)^{n - j + 1}{n \choose j}p^{x + n - j}\cL \mod \pi.
\end{eqnarray}

Putting equations (<ref>) and (<ref>) in equation (<ref>) and using Lemma <ref>, we get

\begin{eqnarray*}
g_1(z) & \equiv & \sum_{a = 1}^{p - 1}\left[\sum_{j = 0}^{n - 1}{n \choose j}p^{1 + x}a^{n - j - 1}(z - a)^j\right]\mathbbm{1}_{a + p\Zp} + \left[\sum_{j = 0}^{n - 1}(-1)^{n - j + 1}{n \choose j}p^{x + n - j}\cL z^j\right]\mathbbm{1}_{p\Zp} \\
& \equiv & \sum_{a = 1}^{p - 1}p^{1 + x}a^{-1}\left[z^n - (z - a)^n\right]\mathbbm{1}_{a + p\Zp} + \left[\sum_{j = 0}^{n - 1}(-1)^{n - j + 1}{n \choose j}p^{x + n - j}\cL z^j\right]\mathbbm{1}_{p\Zp} \\
&& \qquad \qquad \mod \pi \latticeL{k}.
\end{eqnarray*}

Using Lemma <ref> for the first sum and Lemma <ref> for the second sum above, we get

\[
g_1(z) \equiv \sum_{a = 1}^{p - 1}p^{1 + x}a^{-1}z^n\mathbbm{1}_{a + p\Zp} + \left[\sum_{j = \lceil r/2 \rceil}^{n - 1}(-1)^{n - j + 1}{n \choose j}p^{x + n - j}\cL z^j\right]\mathbbm{1}_{p\Zp} \mod \pi \latticeL{k}.
\]

This proves the proposition since $g(z) \equiv g_1(z) \!\! \mod \pi \latticeL{k}.$

Let $i = 1, \> 2, \> \ldots, \> \lceil r/2 \rceil - 1$. If $\nu \leq i - r/2$, then the map $\IZind a^i d^{r - i} \twoheadrightarrow F_{2i, \> 2i + 1}$ factors as

\[
\IZind a^i d^{r - i} \twoheadrightarrow \frac{\IZind a^i d^{r - i}}{\im(\lambda_i^{-1} T_{1, 2} - 1)} \twoheadrightarrow F_{2i, \> 2i + 1},
\]

where $\lambda_i = (-1)^i i {r - i + 1 \choose i}p^{r/2 - i}\cL$. Moreover for $\nu = i - r/2$, the second map in the display above induces a surjection

\[
\pi(r - 2i, \lambda_i, \omega^i) \twoheadrightarrow F_{2i, \> 2i + 1}.
\]

All congruences in this proof are in the space $\br{\latticeL{k}}$ modulo the image of the subspace $\IZind \oplus_{j < r - i}\Fq X^{r - j}Y^j$ under $\IZind \SymF{k - 2} \twoheadrightarrow \br{\latticeL{k}}$.

Fix an $i = 1, 2, \ldots, \lceil r/2 \rceil - 1$. Applying Proposition <ref> with $n = r - i + 1$ and the remark above, we get

\begin{eqnarray}\label{Main equation in leq}
0 \equiv p^{1 + x}\sum_{a = 1}^{p - 1}a^{-1}z^{r - i + 1}\mathbbm{1}_{a + p\Zp} + (r - i + 1)p^{x + 1}\cL z^{r - i}\mathbbm{1}_{p\Zp}.
\end{eqnarray}

First assume that $\nu < i - r/2$. Then $x > -1$. So write

\[
p^{1 + x}\sum_{a = 1}^{p - 1}a^{-1}z^{r - i + 1}\mathbbm{1}_{a + p\Zp} = \sum_{a = 1}^{p - 1}\sum_{j = 0}^{r - i + 1}p^{1 + x}a^{-1}{r - i + 1 \choose j}a^{r - i + 1 - j}(z - a)^{j}\mathbbm{1}_{a + p\Zp}.
\]

Using Lemma <ref>, we see that the terms on the right side of the equation above are congruent to $0$ modulo $\pi \latticeL{k}$. Therefore we get

\[
0 \equiv (r - i + 1)p^{1 + x}\cL z^{r - i}\mathbbm{1}_{p\Zp}.
\]

Applying $\beta = \begin{pmatrix}0 & 1 \\ p & 0\end{pmatrix}$, we get

\[
0 \equiv -(r - i + 1)p^{1 + x + r/2 - i}\cL z^i \mathbbm{1}_{p\Zp}.
\]

Since $x + \nu = r/2 - n = i - 1 - r/2$ and $1 \leq r - i + 1 \leq p - 1$, we see that the coefficient $-(r - i+1)p^{1 + x + r/2 - i}\cL$ is a $p$-adic unit. Cancelling this coefficient, we get

\begin{eqnarray}\label{nu strictly less than i - r/2}
0 \equiv z^i\mathbbm{1}_{p\Zp}.
\end{eqnarray}

This shows that if $\nu < i - r/2$, then the map $\IZind a^i d^{r - i} \twoheadrightarrow F_{2i, \> 2i + 1}$ factors as

\[
\IZind a^i d^{r - i} \twoheadrightarrow \frac{\IZind a^i d^{r - i}}{1} \twoheadrightarrow F_{2i,\> 2i+1}.
\]

Therefore to prove the theorem, we must show that $\lambda_i^{-1} \equiv 0 \mod \pi$. This is immediate since the valuation of $\cL$ is strictly less than $i - r/2$.

Now assume that $\nu = i - r/2$. Therefore $x = -1$.

Applying $\beta$

to (<ref>), we get

\[
0 \equiv \sum_{a = 1}^{p - 1}p^{r/2 - i + 1}a^{-1}z^{i - 1}\mathbbm{1}_{a^{-1}p + p^2\Zp} - (r - i + 1)p^{r/2 - i}\cL z^i \mathbbm{1}_{p\Zp}.
\]

Expanding $z^{i - 1} = (z - a^{-1}p + a^{-1}p)^{i - 1}$ and replacing $a^{-1}$ by $a$, we get

\[
0 \equiv \sum_{a = 1}^{p - 1}\sum_{j = 0}^{i - 1} {i - 1 \choose j}p^{r/2 - i}(ap)^{i - j}(z - ap)^{j} \mathbbm{1}_{ap + p^2\Zp} - (r - i + 1)p^{r/2 - i}\cL z^i\mathbbm{1}_{p\Zp}.
\]

Since $i - j > 0$, we may also include the $a = 0$ term in the first sum. So write

\begin{eqnarray}\label{Main congruence in leq in the equality case}
0 \equiv \sum_{a = 0}^{p - 1}\sum_{j = 0}^{i - 1} {i - 1 \choose j}p^{r/2 - i}(ap)^{i - j}(z - ap)^{j} \mathbbm{1}_{ap + p^2\Zp} - (r - i + 1)p^{r/2 - i}\cL z^i\mathbbm{1}_{p\Zp}.
\end{eqnarray}

Next we note that $\nu = i - r/2$ implies that $\nu > 1 - r/2, \> 2 - r/2, \> \ldots, \> (i - 1) - r/2$. Therefore using Proposition <ref> and the remark in the first paragraph of the proof, we get

\[
z^{r - n}\mathbbm{1}_{p\Zp} \equiv \sum_{b = 1}^{p - 1}\sum_{l = 1}^{i}c(n, l)b^{r - l - n}(z - b)^{l} \mathbbm{1}_{b + p\Zp}.
\]

Applying $\sum\limits_{a = 0}^{p - 1}a^{i - r + n}\begin{pmatrix}1 & 0 \\ -pa & p\end{pmatrix}$, we get

\begin{eqnarray*}
&& \sum_{a = 0}^{p - 1}p^{n - r/2}a^{i - r + n}(z - ap)^{r - n}\mathbbm{1}_{ap + p^2 \Zp} \\
&& \equiv \sum_{a = 0}^{p - 1}\sum_{b = 1}^{p - 1}\sum_{l = 1}^{i} p^{r/2 - l}c(n, l)a^{i - r + n}b^{r - l - n}(z - (a + b)p)^l\mathbbm{1}_{(a + b)p + p^2\Zp}.
\end{eqnarray*}

Making the substitution $a + b = \lambda$ and using

\[
\sum_{b = 1}^{p - 1}(\lambda - b)^{i - r + n}b^{r - l - n} \equiv
\begin{cases}
(-1)^{l - r + n + 1}{i - r + n \choose l - r + n}\lambda^{i - l} & \text{ if } l \geq r - n, \\
0 & \text{ if } l < r - n
\end{cases} \mod p
\]

and Lemma <ref>, we get

\begin{eqnarray*}
&& \sum_{a = 0}^{p - 1}p^{n - r/2}a^{i - r + n}(z - ap)^{r - n}\mathbbm{1}_{ap + p^2\Zp} \\
&& \equiv \sum_{\lambda = 0}^{p - 1}\sum_{l = r - n}^{i}p^{r/2 - l}(-1)^{l - r + n + 1}\lambda^{i - l}{i - r + n \choose l - r + n}c(n, l)(z - \lambda p)^l\mathbbm{1}_{\lambda p + p^2\Zp}.
\end{eqnarray*}

Renaming $\lambda$ as $a$ and plugging the value of $c(n, l)$ obtained in Proposition <ref>, we get one equation

\begin{eqnarray}\label{Refined inductive steps in leq}
0 \equiv \sum_{a = 0}^{p - 1}\sum_{l = r - n}^{i}{i - r + n \choose l - r + n}\frac{(r - n)! n}{(r - l) \cdots (n - l)}p^{r/2 - l}a^{i - l}(z - ap)^{l}\mathbbm{1}_{ap + p^2\Zp},
\end{eqnarray}

for each $n = r, \> r - 1, \> \ldots, \> r - i + 2$.

Finally note that $p^{r/2 - i}z^i\mathbbm{1}_{p\Zp} \equiv 0$ since $r/2 - i > 0$. Therefore expanding $z^i = (z - ap + ap)^i$, we get

\begin{equation}\label{Trivial equation in leq}
0 \equiv\sum_{j = 0}^{i}p^{r/2 - i}{i \choose j}(ap)^{i - j}(z - ap)^j\mathbbm{1}_{ap + p^2\Zp}.
\end{equation}

Using equations (<ref>), (<ref>) and (<ref>), we wish to obtain terms in $(z - ap)$ involving only the $i^{\mathrm{th}}$ power. To this end, consider the matrix equation

\begin{eqnarray}\label{Matrix equation in leq}
\begin{pmatrix}
{i \choose 0}\frac{r}{r} & 0 & 0 & \cdots & {i \choose 0} & {i - 1 \choose 0} \\
{i \choose 1}\frac{r}{r - 1} & {i - 1 \choose 0}\frac{r - 1}{(r - 1)(r - 2)} & 0 & \cdots & {i \choose 1} & {i - 1 \choose 1} \\
{i \choose 2}\frac{r}{r - 2} & {i - 1 \choose 1}\frac{r - 1}{(r - 2)(r - 3)} & {i - 2 \choose 0}\frac{2!(r - 2)}{(r - 2)(r - 3)(r - 4)} & \cdots & {i \choose 2} & {i - 1 \choose 2} \\
\vdots & & & & \vdots & \vdots \\
{i \choose i}\frac{r}{r - i} & {i - 1 \choose i - 1}\frac{r - 1}{(r - i)(r - i - 1)} & {i - 2 \choose i - 2}\frac{2!(r - 2)}{(r - i)(r - i - 1)(r - i - 2)} & \cdots & {i \choose i} & 0
\end{pmatrix}
\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_i \\ x_{i + 1}\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{pmatrix}.
\end{eqnarray}

Once we have the solution to this matrix equation, we multiply the $n = r, \> r - 1, \> r - 2, \> \ldots, \> r - i + 2$ instances of equation (<ref>) by $x_1, \> x_2, \> x_3, \> \ldots, \> x_{i - 1}$, equation (<ref>) by $x_i$, equation (<ref>) by $x_{i + 1}$ and add them all to get

\begin{eqnarray}\label{Crude final congruence in leq}
0 \equiv \sum_{a = 0}^{p - 1}p^{r/2 - i}(z - ap)^{i}\mathbbm{1}_{ap + p^2\Zp} - x_{i + 1}(r - i + 1)p^{r/2 - i}\cL z^i\mathbbm{1}_{p\Zp}.
\end{eqnarray}

We first find the value of $x_{i + 1}$. We again use Cramer's rule. Let $A$ be the $(i + 1)\times (i + 1)$ matrix on the left side of equation (<ref>). Let $A_{i + 1}$ be the same matrix, but with the last column replaced by the column vector on the right side of equation (<ref>). Then

\[
x_{i + 1} = \frac{\det A_{i + 1}}{\det A}.
\]

First consider the matrix $A$. Recall that we number the rows and columns starting with $0$. Perform the operation $C_{i - 1} \to C_{i - 1} - C_0$ to make the $(0, i - 1)^{\mathrm{th}}$ entry $0$. Then for $j \geq 1$, the $(j, i - 1)^{\mathrm{th}}$ entry in the resulting matrix is

\[
{i \choose j} - {i \choose j}\frac{r}{r - j} = -{i - 1 \choose j - 1}\frac{i}{r - j}.
\]

Next, perform the operation $C_{i - 1} \to C_{i - 1} + (r - 2)\frac{i}{r - 1}C_1$ to make the $(1, i - 1)^{\mathrm{th}}$ entry $0$. Then for $j \geq 2$, the $(j, i - 1)^{\mathrm{th}}$ entry in the resulting matrix is

\[
-{i - 1 \choose j - 1}\frac{i}{r - j} + (r - 2)\frac{i}{r - 1}{i - 1 \choose j - 1}\frac{r - 1}{(r - j)(r - j - 1)} = {i - 2 \choose j - 2}\frac{i(i - 1)}{(r - j)(r - j - 1)}.
\]

Continuing this process, after making $(i - 1)$ entries in the second-to-last column $0$, the $(j, i - 1)^{\mathrm{th}}$ entry for $j \geq i - 1$ is

\[
(-1)^{i - 1}{1 \choose j - i + 1}\frac{i(i - 1)\cdots 2}{(r - j)(r - j - 1) \cdots (r - j - i + 2)}.
\]

Now we reduce the last column. Perform the operation $C_i \to C_i - C_0$ to make the $(0, i)^{\mathrm{th}}$ entry $0$. Then for $j \geq 1$, the $(j, i)^{\mathrm{th}}$ entry in the resulting matrix is

\[
{i - 1 \choose j} - {i \choose j}\frac{r}{r - j} = -{i - 1 \choose j - 1}\frac{r + i - j}{r - j}.
\]

Next, perform the operation $C_i \to C_i + \frac{(r - 2)(r + i - 1)}{r - 1}C_1$ to make the $(1, i)^{\mathrm{th}}$ entry $0$. Then for $j \geq 2$, the $(j, i)^{\mathrm{th}}$ entry in the resulting matrix is

\begin{eqnarray*}
&& -{i - 1 \choose j - 1}\frac{r + i - j}{r - j} + \frac{(r - 2)(r + i - 1)}{r - 1}{i - 1 \choose j - 1}\frac{r - 1}{(r - j)(r - j - 1)} \\
&& = {i - 2 \choose j - 2}\frac{(i - 1)(2(r - 1) + i - j)}{(r - j)(r - j - 1)}.
\end{eqnarray*}

Continuing this process, after making $i - 1$ entries in the last column $0$, the $(j, i)^{\mathrm{th}}$ entry for $j \geq i - 1$ in the resulting matrix is

\[
(-1)^{i}{1 \choose j - i + 1}\frac{(i - 1)![(i - 1)(r - i + 2) + (i - j)]}{(r - j)(r - j - 1)\cdots (r - j - i + 2)}.
\]

Now, the reduction for the second-to-last column in the matrix $A_{i + 1}$ is the same as that in the matrix $A$. Moreover, the last column in the matrix $A_{i + 1}$ has $i$ zeros and $1$ in the last entry. Therefore the ratio

\begin{eqnarray*}
\frac{\det A}{\det A_{i + 1}} & = & (-1)^{i}\left[\frac{(i - 1)![(i - 1)(r - i + 2)]}{(r - i)\cdots (r - 2i + 2)} - \frac{(i - 1)![(i - 1)(r - i + 2) + 1]}{(r - i)\cdots (r - 2i + 2)}\right] \\
& = & (-1)^{i}\frac{(i - 1)!}{(r - i)\cdots (r - 2i + 2)}.
\end{eqnarray*}

\[
x_{i + 1} = (-1)^{i}\frac{(r - i)\cdots (r - 2i + 2)}{(i - 1)!}.
\]

Substituting this in equation (<ref>), we get

\[
0 \equiv \sum_{a = 0}^{p - 1}p^{r/2 - i}(z - ap)^i\mathbbm{1}_{ap + p^2\Zp} - (-1)^{i}i {r - i + 1 \choose i}p^{r/2 - i}\cL z^i\mathbbm{1}_{p\Zp}.
\]

This equation is the image of $(T_{1, 2} - \lambda_i)\llbracket \id, X^iY^{r - i}\rrbracket$ under $\IZind a^i d^{r - i} \twoheadrightarrow F_{2i, \> 2i + 1}$. Indeed, by

\[
T_{1, 2}\llbracket \id, X^iY^{r - i}\rrbracket = \sum_{\lambda \in I_1}\left\llbracket \begin{pmatrix}1 & 0 \\ -p\lambda & p\end{pmatrix}, X^iY^{r - i}\right\rrbracket
\mapsto \sum_{\lambda \in I_1}p^{r/2 - i}(z - \lambda p)^i\mathbbm{1}_{\lambda p + p^2 \Zp}.
\]

Therefore we have proved that the surjection $\IZind a^{i}d^{r - i} \twoheadrightarrow F_{2i, \> 2i + 1}$ factors as

\[
\IZind a^{i}d^{r - i} \twoheadrightarrow \frac{\IZind a^{i}d^{r - i}}{\im(\lambda_i^{-1}T_{1, 2} - 1)} \twoheadrightarrow F_{2i, \> 2i + 1}.
\]

Consider the following exact sequences

\[
\begin{tikzcd}
0 \arrow[r] & \im T_{1, 2} \arrow[r] \arrow[d, two heads] & \IZind a^{i}d^{r - i} \arrow[r]\arrow[d, two heads] & \dfrac{\IZind a^{i}d^{r - i}}{\im T_{1, 2}} \arrow[r]\arrow[d, two heads] & 0 \\
0 \arrow[r] & S \arrow[r] & F_{2i, \> 2i + 1} \arrow[r] & Q \arrow[r] & 0,
\end{tikzcd}
\]

where $S$ is the image of $\im T_{1, 2}$ under the surjection $\IZind a^id^{r - i} \twoheadrightarrow F_{2i, \> 2i + 1}$ and $Q = F_{2i, \> 2i + 1}/S$. Since $\im (\lambda_i^{-1}T_{1, 2} - 1)$ maps to $0$ under the middle vertical map in the diagram above, we see that the right vertical map is $0$. Therefore we have a surjection $\im T_{1, 2} \twoheadrightarrow F_{2i, \> 2i + 1}$. As before,
by the first isomorphism theorem and <cit.>, we get a surjection

\[
\frac{\IZind a^{i}d^{r - i}}{\im T_{-1, 0}} \twoheadrightarrow F_{2i, \> 2i + 1},
\]

which factors through

\[
\frac{\IZind a^{i}d^{r - i}}{\im T_{-1, 0} + \im (\lambda_i^{-1}T_{1, 2} - 1)}
% =  \frac{\IZind a^{i}d^{r - i}}{\im T_{-1, 0} + \im (T_{1, 2} - \lambda_i)}
\twoheadrightarrow F_{2i, \> 2i + 1}.
\]

The space on the left is

\[
\frac{\IZind a^{i}d^{r - i}}{\im T_{-1, 0} + \im (T_{1, 2} - \lambda_i)} = \pi(r - 2i, \lambda_i, \omega^{i}).
\]

Therefore we have proved the theorem.

§ ANALYSIS OF $\BR{\LATTICEL{K}}$ AROUND THE LAST POINT FOR ODD WEIGHTS

In the previous section, we gave a uniform description of $\br{\latticeL{k}}$ around all points appearing in Theorem <ref> but the last point, namely, $\nu = 0$ for even $k$ and $\nu = 0.5$ for odd $k$. These two points demand separate treatments.

In this section, we assume that $k$ and $r$ are odd, and describe the behavior of $\br{\latticeL{k}}$ around the point $\nu = 0.5$. The self-dual case of the mod $p$ local Langlands correspondence arises, making the analysis trickier.

§.§ The $\nu \geq 0.5$ case for odd weights

In this section, we prove that if $\nu \geq 0.5$, then the map $\IZind a^{\frac{r-1}{2}}d^{\frac{r+1}{2}} \twoheadrightarrow F_{r-1, \> r}$ factors as

\[
\IZind a^{\frac{r-1}{2}}d^{\frac{r+1}{2}} \twoheadrightarrow \frac{\IZind a^{\frac{r-1}{2}}d^{\frac{r+1}{2}}}{\im(T_{-1, 0}^2 - c T_{-1, 0} + 1)} \twoheadrightarrow F_{r-1, \> r},
\]

\[
c = (-1)^{\frac{r+1}{2}}\dfrac{r+1}{2}\left(\dfrac{\cL - H_{-} - H_{+}}{p^{0.5}}\right).
\]

Moreover, we claim that for $i = \dfrac{r+2}{2}$ there is a surjection

\[
\pi(p - 2, \lambda_i, \omega^{\frac{r + 1}{2}}) \oplus \pi(p - 2, \lambda_i^{-1}, \omega^{\frac{r + 1}{2}}) \twoheadrightarrow F_{r - 1, \> r},
\]

where $\lambda_i + \lambda_i^{-1} = c.$

To prove this, we need inductive steps similar to equation (<ref>), but with an extra $\sqrt{p}$ in the denominator. This is provided by Proposition <ref>. Let us start by proving the following lemmas.  We remark that part 3.
of the next two lemmas will only be used in Section <ref>.

Let $\dfrac{r + 1}{2} \leq n \leq r \leq p - 2$ and $z_0 \in \Zp$. Fix $x \in \bQ$. Define

\[
g(z) = p^x(z - z_0)^n\logL(z - z_0)\mathbbm{1}_{\Zp}.
\]

For $0 \leq a \leq p^{h - 1} - 1$, $0 \leq \alpha \leq p - 1$ and $0 \leq j \leq n - 1$, if

* $\nu > 0, \> \dfrac{r + 3}{2} \leq n \leq r$, $x \geq - 1.5$ and $h \geq 3$, or

* $ \nu > 0, \> n = \dfrac{r + 1}{2}$, $x \geq -1$ and $h \geq 4$, or

* $-0.5 < \nu < 0.5, \> n = \dfrac{r + 1}{2}$, $x + \nu = -0.5$ and $h \geq 3$,

then we have

\[
g^{(j)}(a + \alpha p^{h - 1}) \equiv g^{(j)}(a) + \alpha p^{h - 1}g^{(j + 1)}(a) + \cdots + \frac{(\alpha p^{h - 1})^{n - 1 - j}}{(n - 1 - j)!}g^{(n - 1)}(a) \mod (p^{h - 1})^{r/2 - j}\pi.
\]

Moreover, we have

\begin{eqnarray*}
&& g^{(j)}(a + \alpha p^{h - 1})(z - a - \alpha p^{h - 1})^j\mathbbm{1}_{a + \alpha p^{h - 1} + p^h\Zp} \\
&& \equiv \left[g^{(j)}(a) + \alpha p^{h - 1}g^{(j + 1)}(a) + \cdots + \frac{(\alpha p^{h - 1})^{n - 1 - j}}{(n - 1 - j)!}g^{(n - 1)}(a)\right](z - a - \alpha p^{h - 1})^{j}\mathbbm{1}_{a + \alpha p^{h - 1} + p^h \Zp}
\end{eqnarray*}

modulo $\pi \latticeL{k}$.

The proof of this lemma is analogous to the proof of Lemma <ref>. Consider two cases.

* $v_p(a - z_0) < h - 1$: Recall equation (<ref>) from Lemma <ref>:

\begin{eqnarray}\label{Far away from z0 in Truncated Taylor expansion in self-dual}
g^{(j)}(a + \alpha p^{h - 1}) & = & g^{(j)}(a) + \cdots + \frac{(\alpha p^{h - 1})^{n - 1 - j}}{(n - 1 - j)!}g^{(n - 1)}(a) \nonumber\\
& & + \frac{(\alpha p^{h - 1})^{n - j}}{(n - j)!}n!p^{x}[\logL(a - z_0) + H_n] \nonumber\\
& & + \sum_{l \geq 1}\frac{(\alpha p^{h - 1})^{n - j + l}}{(n - j + l)!}n! p^{x} \frac{(-1)^{l - 1}}{(a - z_0)^l}(l - 1)!.
\end{eqnarray}

Using $\nu > 0$, we see that the valuation of the term in the second line
on the right side of the equation above is greater than or equal to $(n - j)(h - 1) + x$ and using $-0.5 < \nu < 0.5$ and $x + \nu = -0.5$, it is strictly greater than $(n - j)(h - 1) - 1$. The first estimate is strictly greater than $(h - 1)(r/2 - j)$ in the first two cases stated in this lemma. The second estimate is greater than or equal to $(h - 1)(r/2 - j)$ in the third case.

Next, the valuation of a term in the third line
on the right side of equation (<ref>) is greater than or equal to $x + (n - j)(h - 1) + l - \log_p(n - j + l)$. This estimate is again strictly greater than $(h - 1)(r/2 - j)$ in all the cases stated in the lemma.

* $v_p(a - z_0) \geq h - 1$: Using the derivative formula (<ref>) for $g(z)$, we see that the valuations of $g^{(j)}(a + \alpha p^{h - 1}), \> g^{(j)}(a), \>\ldots, (\alpha p^{h - 1})^{n - 1 - j}g^{(n - 1)}(a)$ are greater than or equal to $(n - j)(h - 1) + x$ in the first two cases and strictly greater than $(n - j)(h - 1) - 1$ in the third case. The first estimate is strictly greater than $(h - 1)(r/2 - j)$ in the first two cases stated in the lemma as seen above. The second estimate is greater than or equal to $(h - 1)(r/2 - j)$ in the third case again as seen above.

The final claim made in this lemma is a consequence of the first claim made in this lemma and Lemma <ref>.

Let $\dfrac{r + 1}{2} \leq n \leq r \leq p - 2$ and $z_0 \in \Zp$. Fix $x \in \bQ$. Then for

\[
g(z) = p^x(z - z_0)^n\logL(z - z_0)\mathbbm{1}_{\Zp},
\]

we have

* $g(z) \equiv g_2(z) \mod \pi \latticeL{k}$ if $\nu > 0, \> \dfrac{r + 3}{2} \leq n \leq r$ and $x \geq -1.5$,

* $g(z) \equiv g_3(z) \mod \pi \latticeL{k}$ if $\nu > 0, \> n = \dfrac{r + 1}{2}$ and $x \geq -1$, or

* $g(z) \equiv g_2(z) \mod \pi \latticeL{k}$ if $-0.5 < \nu < 0.5$, $n = \dfrac{r + 1}{2}$ and $x + \nu = -0.5$.

The proof of this lemma is exactly the same as that of Lemma <ref> but with Lemma <ref> replaced by Lemma <ref>.

The following lemma is used in the proof of the next lemma.

Let $\dfrac{r + 1}{2} \leq n \leq r \leq p - 2$ and $z_0 \in \Zp$. Fix $x \in \bQ$. Define

\[
f(z) = p^xz^{r - n}(1 - zz_0)^n\logL(1 - zz_0)\mathbbm{1}_{p\Zp}.
\]

For $0 \leq a \leq p^{h - 1} - 1$, $0 \leq \alpha \leq p - 1$ and $0 \leq j \leq n - 1$, if

* $\dfrac{r + 3}{2} \leq n \leq r$, $x \geq -1.5$, and $h \geq 3$, or

* $n = \dfrac{r + 1}{2}$, $x \geq - 1$ and $h \geq 4$,

then we have

\[
f^{(j)}(a + \alpha p^{h - 1}) \equiv f^{(j)}(a) + \alpha p^{h - 1}f^{(j + 1)}(a) + \cdots + \frac{(\alpha p^{h - 1})}{(n - 1 - j)!}f^{(n - 1)}(a) \mod (p^{h - 1})^{r/2 - j}\pi.
\]

Moreover, we have

\begin{eqnarray*}
&& f^{(j)}(a + \alpha p^{h - 1})(z - a - \alpha p^{h - 1})^j\mathbbm{1}_{a + \alpha p^{h - 1}+ p^h\Zp} \\
&& \equiv \left[f^{(j)}(a) + \alpha p^{h - 1}f^{(j + 1)}(a) + \cdots + \frac{(\alpha p^{h - 1})}{(n - 1 - j)!}f^{(n - 1)}(a)\right](z - a - \alpha p^{h - 1})^j\mathbbm{1}_{a + \alpha p^{h - 1} + p^h \Zp}
\end{eqnarray*}

modulo $\pi \latticeL{k}$.

Since $f(z)$ has $\mathbbm{1}_{p\Zp}$ as a factor, this lemma is vacuously true if $p \nmid a$. So assume that $p \mid a$. The $j^{\mathrm{th}}$ derivative of $f$ is given in (<ref>). We expand $f^{(j)}(a + \alpha p^{h - 1})$ as a Taylor series:

\begin{eqnarray*}
f^{(j)}(a + \alpha p^{h - 1}) & = & f^{(j)}(a) + \cdots + \frac{(\alpha p^{h - 1})^{n - 1 - j}}{(n - 1 - j)}f^{(n - 1)}(a) \\
&& + \sum_{l \geq 1}\frac{(\alpha p^{h - 1})^{n - 1 - j + l}}{(n - 1 - j + l)!}f^{(n - 1 + l)}(a).
\end{eqnarray*}

Using the derivative formula (<ref>), we see that the valuation of the $l^{\mathrm{th}}$ summand in the second line above is greater than or equal to $x + (h - 1)(n - 1 - j + l) - v_p((n - 1 - j + l)!)$.

* Note that $-1.5 + (h - 1)(n - 1 - j + l) - v_p((n - 1 - j + l)!) > (h - 1)(r/2 - j)$ is true for any $h \geq 3$ if and only if it is true for $h = 3$. We check this inequality for $h = 3$. Indeed, the inequality is clearly true using the estimate $v_p(t!) \leq \frac{t}{p - 1}.$

* Note that $-1 + (h - 1)(n - 1 - j + l) - v_p((n - 1 - j + l)!)$ is not greater than $(h - 1)(r/2 - j)$ for $h = 3$ and $l = 1$. Therefore we have to take $h \geq 4$. We then check the inequality $-1 + (h - 1)(n - 1 - j + l) - v_p((n - 1 - j + l)!) > (h - 1)(r/2 - j)$ for $h = 4$. Indeed, this is clear for $l = 1$ and for $l \geq 2$, we see this again using the estimate $v_p(t!) \leq \frac{t}{p - 1}$.

The second claim in this lemma follows from the first claim by using Lemma <ref>.

The following lemma is used to show that $g(z) \mathbbm{1}_{\qp \setminus \zp} \equiv 0 \mod \pi \latticeL{k}$ for the functions $g(z)$ appearing (twice in) Proposition <ref>, Proposition <ref>, Proposition <ref>.

Let $\dfrac{r + 1}{2} \leq n \leq r \leq p - 2$. For $i$ in a finite indexing set $I$, let $\lambda_i \in E$ and $z_i \in \Zp$ be such that $\sum_{i \in I}\lambda_i z_i^j = 0$ for all $0 \leq j \leq n$. If $n = \dfrac{r + 1}{2}$, then we also assume that $\sum_{i \in I}\lambda_i z_i^{n + 1} = 0$. Fix an $x \in \bQ$. Then for

\[
f(z) = \sum_{i \in I}p^x \lambda_i z^{r - n}(1 - zz_i)^n\logL(1 - zz_i)\mathbbm{1}_{p\Zp},
\]

we have

* $f(z) \equiv 0 \mod \pi \latticeL{k}$ if $\dfrac{r + 3}{2} \leq n \leq r$ and $x \geq -1.5$.

* $f(z) \equiv 0 \mod \pi \latticeL{k}$ if $n = \dfrac{r + 1}{2}$ and $x \geq -1$.

We have

* As in the proof of Lemma <ref> with Lemma <ref> replaced by Lemma <ref>, we see that $f(z) \equiv f_2(z) \mod \pi \latticeL{k}$. We now prove that $f_2(z) \equiv 0 \mod \pi \latticeL{k}$.
This case does not occur if $r = 1$ thanks to the inequality $\dfrac{r + 3}{2} \leq r$. So we assume that $r \geq 3$ (since $r$ is odd). Recall that

\[
f_2(z) = \sum_{a = 0}^{p^2 - 1}\sum_{j = 0}^{n - 1}\frac{f^{(j)}(a)}{j!}(z - a)^j\mathbbm{1}_{a + p^2\Zp}.
\]

Working as in the proof of Lemma <ref>, we see that

\[
v_p\left(\frac{f^{(j)}(\alpha p)}{j!}\right) \geq \min_{l \geq n - m + 1}\{-1.5 + r - n - j + m + l - \lfloor \log_p(l)\rfloor\}.
\]

A short computation shows that $-1.5 + r - n - j + m + l - \lfloor \log_p(l) \rfloor > r/2 - j$ for $l \geq n - m + 1$. Therefore Lemma <ref> implies that $f_2(z) \equiv 0 \mod \pi \latticeL{k}$.

* In this case, $f(z) \equiv f_3(z) \mod \pi \latticeL{k}$ again using Lemma <ref>. We show that $f_3(z) \equiv 0 \mod \pi \latticeL{k}$. Recall that

\[
f_3(z) = \sum_{a = 0}^{p^3 - 1}\sum_{j = 0}^{n - 1}\frac{f^{(j)}(a)}{j!}(z - a)^j\mathbbm{1}_{a + p^3\Zp}.
\]

We similarly see that

\[
v_p\left(\frac{f^{j}(\alpha p)}{j!}\right) \geq \min_{l \geq n - m + 2}\{-1 + r - n - j + m + l - \lfloor \log_p(l)\rfloor\},
\]

A short computation using $p > 3$ shows that $-1 + r - n - j + m + l -\lfloor \log_p(l) \rfloor > 2(r/2 - j)$ for all $l \geq n - m + 2$. So $f_3(z) \equiv 0 \mod \pi \latticeL{k}$ by Lemma <ref>.

Let $\dfrac{r + 3}{2} < n \leq r$ and $0 \leq j \leq \dfrac{r + 1}{2}$. For $0 \leq a, \> i \leq p - 1$, let $a_i \in \Zp$ be such that $a - [i] = [a - i] + pa_i$. Set $\lambda_0 = 1 - p$ and $\lambda_i = 1$ when $1 \leq i \leq p - 1$. Then we have

\[
\sum_{\substack{i = 0 \\ i \neq a}}\frac{\lambda_i}{p^{1.5}}[a - i]^{n - j - 1}pa_i = \frac{a^{n - j}}{p^{0.5}(n - j)} \mod \pi.
\]

The proof is similar to the proof of Lemma <ref>, noting that
$\pi \mid p^{0.5}$.

Since $[0] = 0$ and $n - j - 1 > 0$, we see that

\[
\sum_{\substack{i = 0 \\ i \neq a}}\frac{\lambda_i}{p^{1.5}}[a - i]^{n - j - 1}pa_i = \sum_{i = 0}\frac{\lambda_i}{p^{1.5}}[a - i]^{n - j - 1}pa_i.
\]

\[
\sum_{i = 0}^{p - 1}\frac{\lambda_i}{p^{1.5}}(a - [i])^{n - j} \equiv \sum_{i = 0}^{p - 1}\frac{\lambda_i}{p^{1.5}}\left([a - i]^{n - j} + (n - j)[a - i]^{n - j - 1}pa_i\right) \mod \pi.
\]

Expanding the left side, plugging in the values of $\lambda_i$ and using equation (<ref>), we get

\[
0 \equiv \frac{-[a]^{n - j}}{p^{0.5}} + \sum_{i = 0}^{p - 1}\frac{\lambda_i}{p^{1.5}}(n - j)[a - i]^{n - j - 1}pa_i \mod \pi.
\]

\[
\sum_{i = 0}^{p - 1}\frac{\lambda_i}{p^{1.5}}[a - i]^{n - j - 1}pa_i \equiv \frac{[a]^{n - j}}{p^{0.5}(n - j)} \equiv \frac{a^{n - j}}{p^{0.5}(n - j)} \mod \pi. \qedhere
\]

We now state and prove the following (deep) inductive steps. The word `deep' is meant to signify that the power of $p$ appearing here is $p^{1/2}$ deeper than the power of $p$ appearing in the earlier inductive steps in Proposition <ref> (more precisely, equation (<ref>)).

Let $\nu > 0$. Then for each $0 \leq a \leq p - 1$ and $\dfrac{r + 3}{2} \leq n \leq r$, we have

\[
0 \equiv \sum_{\alpha = 0}^{p - 1}\sum_{l = r-n}^{{\frac{r+1}{2}}}p^{\frac{r-1}{2} - l}{n - \frac{r-1}{2} \choose n - r + l}\frac{(r-n)! n}{(r - l)\cdots (n - l)}\alpha^{\frac{r+1}{2} - l}(z - a - \alpha p)^l \mathbbm{1}_{a + \alpha p + p^2\Zp} \mod \pi \latticeL{k}.
\]

We remark that $\dfrac{r + 3}{2} \leq r$ implies $r \neq 1$.

First assume that $\dfrac{r+3}{2} < n \leq r$. Set

\[
g(z) = \sum_{i = 0}^{p - 1}\frac{\lambda_i}{p^{1.5}}(z - [i])^n\logL(z - [i]),
\]

where $\lambda_0 = 1 - p$ and $\lambda_i = 1$ for $1 \leq i \leq p - 1$. Write $g(z) = g(z)\mathbbm{1}_{\Zp} + g(z)\mathbbm{1}_{\Qp \setminus \Zp}$. Using the identity $\sum_{i = 0}^{p - 1}\lambda_i [i]^j = 0$ for $0 \leq j \leq n \leq p - 2$ proved in Lemma <ref> (1), we see that

\[
w \cdot g(z)\mathbbm{1}_{\Qp \setminus \Zp} = \sum_{i = 0}^{p - 1}\frac{\lambda_i}{p^{1.5}}z^{r - n}(1 - z[i])^n\logL(1 - z[i])\mathbbm{1}_{p\Zp}.
\]

A priori, there is an extra term involving $\logL(z)$ on the right side of the equation above
as in the proof of Proposition <ref>.
However, due to the identity $\sum_{i = 0}^{p - 1}\lambda_i[i]^{n} = 0$, this term becomes $0$. Applying Lemma <ref> (1), we see that this function, hence also
$g(z)\mathbbm{1}_{\qp \setminus\Zp}$,  is $0$ modulo $\pi \latticeL{k}$.

So $g(z) \equiv g(z)\mathbbm{1}_{\Zp} \mod \pi \latticeL{k}$.

Lemma <ref> (1) implies that $g(z)\mathbbm{1}_{\Zp} \equiv g_2(z) \mod \pi \latticeL{k}$. Next, we analyze $g_2(z) - g_1(z)$ and $g_1(z)$ separately.

For $0 \leq j \leq n - 1$, the coefficient of $(z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2 \Zp}$ in $g_2(z) - g_1(z)$ is

\begin{eqnarray}\label{jth summand in g2 - g1 in self-dual}
&& \!\!\!\!\!\!\!\! {n \choose j}\sum_{i = 0}^{p - 1} \frac{\lambda_i}{p^{1.5}}\Big[(a + \alpha p - [i])^{n - j}\logL(a + \alpha p - [i]) -
(a - [i])^{n - j}\logL(a - [i]) \\
&& \!\!\!\!\!\!\!\! - \left. {n - j \choose 1}\alpha p (a - [i])^{n - j - 1}\logL(a - [i]) - \ldots - {n - j \choose n - j - 1}(\alpha p)^{n - j - 1}(a - [i])\logL(a - [i])\right]. \nonumber
\end{eqnarray}

First assume that $i \neq a$. Writing $\logL(a + \alpha p - [i]) = \logL(1 + (a - [i])^{-1}\alpha p) + \logL(a - [i])$, we see that the $i^{\mathrm{th}}$ summand in the display above is

\[
\frac{\lambda_i}{p^{1.5}}[(a - [i] + \alpha p)^{n - j}\logL(1 + (a - [i])^{-1}\alpha p) + (\alpha p)^{n - j}\logL(a - [i])].
\]

Since $0 \leq a \leq p - 1$, the condition $i \neq a$ implies that $i \not \equiv a \!\! \mod p$. Therefore the valuation of $\logL(a - [i])$ is greater than or equal to $1$. Since $n > \dfrac{r+3}{2} > \dfrac{r + 1}{2}$, the second term in the display above is congruent to $0$ modulo $p^{r/2 - j}\pi$. Moreover, expanding the first term, we see that the $i \neq a$ summand in expression (<ref>) is

\begin{eqnarray}\label{i not equal to a in self-dual}
\frac{\lambda_i}{p^{1.5}}[(a - [i])^{n - j - 1}\alpha p + c_{n - j - 2}(a - [i])^{n - j - 2}(\alpha p)^2 + \cdots + c_0(\alpha p)^{n - j}] \mod p^{r/2 - j}\pi,
\end{eqnarray}

where $c_{n - j - 2}, \ldots, c_0 \in \Zp$.

Next assume that $i = a$. Using $\nu > 0$, we see that each summand in expression (<ref>) has valuation greater than or equal to $n - j - 1.5$. Since $n > \dfrac{r + 3}{2}$, we see that the $i = a$ summand in expression (<ref>) is $0$ modulo $p^{r/2 - j}\pi$.

Using (<ref>) and the identity $\sum_{i = 0}^{p - 1}\lambda_i(z - [i])^{n - 1} = 0$, we see that expression (<ref>) becomes

\begin{eqnarray}\label{Final expression for the jth term in g2 - g1 in the self-dual case}
-\frac{\lambda_a}{p^{1.5}}[(a - [a])^{n - j - 1}(\alpha p) + \cdots + c_0(\alpha p)^{n - j}] \mod p^{r/2 - j}\pi.
\end{eqnarray}

Each term in (<ref>) has valuation greater than or equal to $n - j - 1.5$. Since $n > \dfrac{r + 3}{2}$, the display above is congruent to $0$ modulo $p^{r/2 - j}\pi$. Therefore

$  g_2(z) - g_1(z) \equiv 0 \mod \pi \latticeL{k}$.

Next we analyze

\begin{eqnarray}\label{g1 in inductive steps in self-dual}
g_1(z) = \sum_{a = 0}^{p - 1}\left[\sum_{j = 0}^{n - 1}\frac{g^{(j)}(a)}{j!}(z - a)^j\right]\mathbbm{1}_{a + p\Zp}.
\end{eqnarray}

Note that

\begin{eqnarray}\label{jth summand in g1 in inductive steps in self-dual}
\frac{g^{(j)}(a)}{j!} = {n \choose j}\sum_{i = 0}^{p - 1}\frac{\lambda_i}{p^{1.5}}(a - [i])^{n - j}\logL(a - [i]).
\end{eqnarray}

First assume that $j > \dfrac{r + 1}{2}$. For $i \neq a$, the $i^{\mathrm{th}}$ summand on the right side of equation (<ref>) has valuation greater than or equal to $-0.5$.
Using Lemma <ref>, we see that the $j > \dfrac{r + 1}{2}$ and $i \neq a$ terms do not contribute to $g_1(z)$. Next, the valuation of the $i = a$ summand in (<ref>) is greater than or equal to $n - j - 1.5$. Since $n > \dfrac{r + 3}{2}$, Lemma <ref> implies that the $j > \dfrac{r + 1}{2}$ and $i = a$ terms do not contribute to $g_1(z)$ as well.

Now assume that $j \leq \dfrac{r + 1}{2}$. The valuation of the $i = a$ summand on the right side of equation (<ref>) is greater than or equal to $n - j - 1.5$. Since $n > \dfrac{r + 3}{2}$, we see that $n - j - 1.5 > 0$. Therefore using Lemma <ref>, we see that the $i = a$ terms do not contribute to $g_1(z)$. Next, for $i \neq a$ write $a - [i] = [a - i] + pa_i$ for some $a_i \in \Zp$. Then $\logL(a - [i]) = \logL(1 + [a - i]^{-1}pa_i)$ since $[a - i]$ is a root of unity. Therefore the $i \neq a$ summand on the right side of (<ref>) is

\begin{eqnarray}\label{i not equal to a in g1 in self-dual}
\frac{\lambda_i}{p^{1.5}}([a - i] + pa_i)^{n - j}\logL(1 + [a - i]^{-1}pa_i) & = & \frac{\lambda_i}{p^{1.5}}[a - i]^{n - j - 1}pa_i \mod \pi.
\end{eqnarray}

Plugging (<ref>) into (<ref>) and using Lemma <ref>, we get

\[
\frac{g^{(j)}(a)}{j!} \equiv {n \choose j} \frac{a^{n - j}}{p^{0.5}(n - j)} \mod \pi.
\]

\[
g_1(z) \equiv \sum_{a = 0}^{p - 1}\sum_{j = 0}^{\frac{r + 1}{2}}{n \choose j}\frac{a^{n - j}}{p^{0.5}(n - j)}(z - a)^j \mathbbm{1}_{a + p\Zp} \mod \pi \latticeL{k}.
\]

This shows that

\begin{eqnarray*}\label{hard inductive steps for large n}
0 \equiv \sum_{a = 0}^{p - 1} \left[ \sum_{j = 0}^{\frac{r + 1}{2}}{n \choose j}\frac{a^{n - j}}{p^{0.5}(n - j)}(z - a)^j \right] \mathbbm{1}_{a + p\Zp} \mod \pi \latticeL{k}.
\end{eqnarray*}

This is equation (<ref>) in the proof of the `shallow' inductive steps
except that the sum over $j$ goes up to
$\dfrac{r+1}{2}$ and there is an extra denominator $p^{0.5}$.
Proceeding exactly as in the proof of in Proposition <ref>), we get

Now the binomial expansion of $z^n = [(z - a) + a]^n$ gives

\begin{eqnarray}\label{Binomial expansion in hard inductive steps for large n}
\frac{1}{p^{0.5}}z^n\mathbbm{1}_{\Zp} \equiv \sum_{a = 0}^{p - 1}\left[\sum_{j = 0}^{\frac{r + 1}{2}}{n \choose j}\frac{a^{n - j}}{p^{0.5}}(z - a)^j\right]\mathbbm{1}_{a + p\Zp} \mod \pi \latticeL{k}.
\end{eqnarray}

We have used Lemma <ref> to truncate the binomial expansion. Subtracting $n$ times equation (<ref>) from equation (<ref>), we get

\[
\frac{1}{p^{0.5}}z^n\mathbbm{1}_{\Zp} \equiv \sum_{a = 0}^{p - 1}\left[\sum_{j = 1}^{\frac{r + 1}{2}} -\frac{j}{n - j}{n \choose j}\frac{a^{n - j}}{p^{0.5}}(z - a)^j\right]\mathbbm{1}_{a + p\Zp} \mod \pi \latticeL{k}.
\]

Note that we have dropped the $j = 0$ term since the factor $\dfrac{j}{n - j}$ is $0$. Since $j < n$, there is a positive power of $a$ on the right side of the equation above. Therefore we drop the $a = 0$ summand. Applying $-w = -\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$, we get

\[
\frac{1}{p^{0.5}}z^{r - n}\mathbbm{1}_{p\Zp} \equiv \sum_{a = 1}^{p - 1}\left[\sum_{j = 1}^{\frac{r + 1}{2}}\frac{j}{n - j}{n \choose j}\frac{a^{n - j}}{p^{0.5}}z^{r - j}(1 - az)^j\right]\mathbbm{1}_{a^{-1} + p\Zp} \mod \pi \latticeL{k}.
\]

Pulling $-a$ out of $(1 - az)^j$, we get

\[
\frac{1}{p^{0.5}}z^{r - n}\mathbbm{1}_{p\Zp} \equiv \sum_{a = 1}^{p - 1}\left[\sum_{j = 1}^{\frac{r + 1}{2}}\frac{(-1)^jj}{n - j}{n \choose j}\frac{a^n}{p^{0.5}}z^{r - j}(z - a^{-1})^j\right]\mathbbm{1}_{a^{-1} + p\Zp} \mod \pi \latticeL{k}.
\]

Expanding $z^{r - j} = \sum\limits_{l = 0}^{r - j}{r - j \choose l}a^{-l}(z - a^{-1})^{r - j - l}$ and replacing $a^{-1}$ by $b$, we get

\[
\frac{1}{p^{0.5}}z^{r - n}\mathbbm{1}_{p\Zp} \equiv \sum_{b = 1}^{p - 1}\left[\sum_{j = 1}^{\frac{r + 1}{2}} \sum_{l = 0}^{r - j} \frac{(-1)^j j}{n - j}{n \choose j}{r - j \choose l}\frac{b^{l - n}}{p^{0.5}}(z - b)^{r - l}\right]\mathbbm{1}_{b + p\Zp} \mod \pi \latticeL{k}.
\]

Using Lemma <ref>, we drop the $l < \dfrac{r - 1}{2}$ terms to get

\[
\frac{1}{p^{0.5}}z^{r - n}\mathbbm{1}_{p\Zp} \equiv \sum_{b = 1}^{p - 1}\left[\sum_{j = 1}^{\frac{r + 1}{2}}\sum_{l = \frac{r - 1}{2}}^{r - j} \frac{(-1)^j j}{n - j}{n \choose j}{r - j \choose l}\frac{b^{l - n}}{p^{0.5}}(z - b)^{r - l}\right]\mathbbm{1}_{b + p\Zp} \mod \pi \latticeL{k}.
\]

Exchanging the order of the sums inside the brackets, we get

\[
\frac{1}{p^{0.5}}z^{r - n}\mathbbm{1}_{p\Zp} \equiv \sum_{b = 1}^{p - 1}\left[\sum_{l = \frac{r - 1}{2}}^{r - 1}\sum_{j = 1}^{r - l} \frac{(-1)^j j}{n - j}{n \choose j}{r - j \choose l}\frac{b^{l - n}}{p^{0.5}}(z - b)^{r - l}\right]\mathbbm{1}_{b + p\Zp} \mod \pi \latticeL{k}.
\]

Replacing $l$ by $r - l$, we get

\[
\frac{1}{p^{0.5}}z^{r - n}\mathbbm{1}_{p\Zp} \equiv \sum_{b = 1}^{p - 1}\left[\sum_{l = 1}^{\frac{r + 1}{2}}\left(\sum_{j = 1}^{l} \frac{(-1)^j j}{n - j}{n \choose j}{r - j \choose r - l}\right)\frac{b^{r - l - n}}{p^{0.5}}(z - b)^{l}\right]\mathbbm{1}_{b + p\Zp} \mod \pi \latticeL{k}.
\]

We remark that because of the slightly deeper denominator,
Lemma <ref> can
only be used in the intermediate step (<ref>)
to drop the $0 \leq l < \dfrac{r - 1}{2}$ terms there.

The expression inside the parentheses is $c(n, l)$ from Proposition <ref>. Applying the operator $\sum\limits_{\alpha = 0}^{p - 1}\alpha^{n - \frac{r - 1}{2}}\begin{pmatrix}1 & 0 \\ -a - \alpha p & p\end{pmatrix}$, we get

\begin{eqnarray*}
&& \sum_{\alpha = 0}^{p - 1}\alpha^{n - \frac{r - 1}{2}}p^{n - \frac{r + 1}{2}}(z - a - \alpha p)^{r - n}\mathbbm{1}_{a + \alpha p + p^2\Zp} \\
&& \equiv \sum_{\alpha = 0}^{p - 1}\sum_{b = 1}^{p - 1}\sum_{l = 1}^{\frac{r + 1}{2}}\alpha^{n - \frac{r - 1}{2}} p^{\frac{r - 1}{2} - l}c(n, l)b^{r - l - n}(z - a - (\alpha + b)p)^l\mathbbm{1}_{a + (\alpha + b)p + p^2\Zp} \mod \pi \latticeL{k}.
\end{eqnarray*}

Making the substitution $\alpha + b = \lambda$ and summing over $b$ using (<ref>) (but with $i = \lceil {r/2} \rceil$), we get

\begin{eqnarray*}
&& \!\!\!\!\!\!\!\! \sum_{\lambda = 0}^{p - 1}\alpha^{n - \frac{r - 1}{2}}p^{n - \frac{r + 1}{2}}(z - a - \alpha p)^{r - n}\mathbbm{1}_{a + \alpha p + p^2\Zp} \\
&& \!\!\!\!\!\!\!\! \equiv \sum_{\lambda = 0}^{p - 1}\sum_{l = r - n}^{\frac{r + 1}{2}}p^{\frac{r - 1}{2} - l}\lambda^{\frac{r + 1}{2} - l}c(n, l)(-1)^{n - r + l + 1}{n - \frac{r - 1}{2} \choose n - r + l}(z - a - \lambda p)^l\mathbbm{1}_{a + \lambda p + p^2\Zp} \!\!\!\!\! \mod \pi \latticeL{k}.
\end{eqnarray*}

Renaming $\lambda$ as $\alpha$ and substituting the value of $c(n, l)$ obtained in
Proposition <ref>,

proves the proposition for $\dfrac{r + 3}{2} < n \leq r$.

Next, assume that $n = \dfrac{r + 3}{2}$. Set

\[
g(z) = \sum_{i = 0}^{p - 1}\frac{\lambda_i}{p^{1.5}}(z - i)^{\frac{r + 3}{2}}\logL(z - i),
\]

where the $\lambda_i$ are as in Lemma <ref> (3), so that
$\lambda_0 = 1$ and $\sum\limits_{i = 0}^{p - 1}\lambda_i i^j = 0$
for $0 \leq j \leq p - 2$. Note that this function is slightly different from the one we chose for $\dfrac{r + 3}{2} < n \leq r$. This change is made to eliminate the terms analogous to the $(a-[a])$ terms in equation (<ref>). We have to make this change since the bound $n - j - 1.5 > r/2 - j$ is no longer true for $n = \dfrac{r + 3}{2}$.

We write $g(z) = g(z)\mathbbm{1}_{\Zp} + g(z)\mathbbm{1}_{\Qp \setminus \Zp}$. Note that

\[
w \cdot g(z)\mathbbm{1}_{\Qp \setminus \Zp}  = \sum_{i = 0}^{p - 1}\frac{\lambda_i}{p^{1.5}}z^{\frac{r - 3}{2}}(1 - zi)^{\frac{r + 3}{2}}\logL(1 - zi)\mathbbm{1}_{p\Zp}.
\]

Applying Lemma <ref> (1), we see
that $g(z)\mathbbm{1}_{\Qp \setminus \Zp} \equiv 0 \mod \pi \latticeL{k}$. Therefore as before, $g(z) \equiv g(z)\mathbbm{1}_{\Zp} \mod \pi \latticeL{k}$. Now Lemma <ref> (1) implies that $g(z)\mathbbm{1}_{\Zp} \equiv g_2(z) \mod \pi \latticeL{k}$. We analyze $g_2(z) - g_1(z)$ and $g_1(z)$ separately.

The coefficient of $(z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2\Zp}$ in $g_2(z) - g_1(z)$ is

\begin{eqnarray}\label{jth summand in g2 - g1 in self-dual for n = r+3/2}
&& \!\!\!\!\!\!\!\!\!\!\!\! {\frac{r + 3}{2} \choose j}\sum_{i = 0}^{p - 1} \frac{\lambda_i}{p^{1.5}}\Big[(a + \alpha p - i)^{\frac{r + 3}{2} - j}\logL(a + \alpha p - i) -
(a - i)^{\frac{r + 3}{2} - j}\logL(a - i) \\
&& \!\!\!\!\!\!\!\!\!\! - \!\!\left. {\frac{r + 3}{2} - j \choose 1}\alpha p (a - i)^{\frac{r + 3}{2} - j - 1}\!\logL(a - i) \! - \! \ldots - \! {\frac{r + 3}{2} - j \choose \frac{r + 3}{2} - j - 1}\!(\alpha p)^{\frac{r + 3}{2} - j - 1}(a - i)\logL(a - i)\right]. \nonumber
\end{eqnarray}

First assume $i \neq a$. Writing $\logL(a + \alpha p - i) = \logL(1 + (a - i)^{-1}\alpha p) + \logL(a - i)$, we see that the $i^{\mathrm{th}}$ summand in the display above is

\[
\frac{\lambda_i}{p^{1.5}}\left[(a - i + \alpha p)^{\frac{r + 3}{2} - j}\logL(1 + (a - i)^{-1}\alpha p) + (\alpha p)^{\frac{r + 3}{2} - j}\logL(a - i)\right].
\]

As before $p \mid \logL(a - i)$. So the second summand above is $0$ modulo $p^{r/2 - j}\pi$. Moreover, expanding the first term, we see that the $i \neq a$ summand in expression (<ref>) is

\begin{eqnarray}\label{i neq a summand in g2 - g1 in self-dual for n = r+3/2}
\frac{\lambda_i}{p^{1.5}}[(a - i)^{\frac{r + 1}{2} - j}\alpha p + c_{\frac{r - 1}{2} - j}(a - i)^{\frac{r - 1}{2} - j}(\alpha p)^2 + \cdots + c_0(\alpha p)^{\frac{r + 3}{2} - j}] \mod p^{r/2 - j}\pi,
\end{eqnarray}

where $c_{\frac{r - 1}{2}-j}, \ldots, c_0 \in \Zp$ with

$c_0 = H_{\frac{r + 3}{2} - j}$, by (<ref>).

Next, assume that $i = a$. Then the $i^{\mathrm{th}}$ summand in expression (<ref>) is

\[
\frac{\lambda_a}{p^{1.5}}(\alpha p)^{\frac{r + 3}{2} - j}\logL(\alpha p) = \frac{\lambda_a}{p^{1.5}}(\alpha p)^{\frac{r + 3}{2} - j}\cL + \frac{\lambda_a}{p^{1.5}}(\alpha p)^{\frac{r + 3}{2} - j}\logL(\alpha).
\]

The second term in the display above is $0$ modulo $p^{r/2 - j}\pi$. This is clear if $\alpha = 0$. If $\alpha \neq 0$, then the valuation of the second term in the right side of the display above is greater than or equal to $(r + 2)/2 - j > r/2 - j$ since $\alpha$ is a $p$-adic unit. So the $i = a$ summand in (<ref>) is

\begin{eqnarray}\label{i = a summand in g2 - g1 in self-dual for n = r+3/2}
\frac{\lambda_a}{p^{1.5}}(\alpha p)^{\frac{r + 3}{2} - j}\cL \mod p^{r/2 - j}\pi.
\end{eqnarray}

Putting (<ref>) and (<ref>) in (<ref>) and using the identity $\sum\limits_{i = 0}^{p - 1}\lambda_i(z - i)^{p - 2} = 0$, we see that the coefficient of $(z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2\Zp}$ in $g_2(z) - g_1(z)$ is

\[
{\frac{r + 3}{2} \choose j}\frac{\lambda_a}{p^{1.5}}(\alpha p)^{\frac{r + 3}{2} - j}(\cL - H_{\frac{r + 3}{2} - j}) \mod p^{r/2 - j}\pi.
\]

Note $\lambda_a \equiv 1 \!\! \mod p$. Also, the
valuation of the coefficient of $(z - a - \alpha p)^{\frac{r + 1}{2}}$ is greater than
or equal to $0.5$, so  we may drop the $j = \dfrac{r + 1}{2}$ term, by
Lemma <ref>. We get

\begin{eqnarray}\label{g2 - g1 in self-dual for n = r+3/2}
g_2(z) - g_1(z) & \equiv & \sum_{a = 0}^{p - 1}\sum_{\alpha = 0}^{p - 1}\left[\sum_{j = 0}^{\frac{r - 1}{2}}{\frac{r + 3}{2} \choose j}\frac{1}{p^{1.5}}(\alpha p)^{\frac{r + 3}{2} - j}\left(\cL - H_{\frac{r + 3}{2} - j}\right)(z - a - \alpha p)^j\right]\mathbbm{1}_{a + \alpha p + p^2\Zp} \nonumber \\
&& \qquad \qquad \mod \pi \latticeL{k}.
\end{eqnarray}

We claim that $g_2(z) - g_1(z) \equiv 0 \!\!\mod \pi \latticeL{k}$. Since $\nu > 0$, Proposition <ref> gives us

\[
z^{r - n}\mathbbm{1}_{p\Zp} \equiv \sum_{b = 1}^{p - 1}\sum_{j = 1}^{\frac{r - 1}{2}}c(n, j)b^{r - j - n}(z - b)^j\mathbbm{1}_{b + p\Zp} \mod \pi \latticeL{k}
\]

for every $\dfrac{r + 3}{2} \leq n \leq r$. Applying $\sum\limits_{\alpha = 0}^{p - 1}\alpha^{n - \frac{r - 3}{2}}\begin{pmatrix}1 & 0 \\ -a - \alpha p & p\end{pmatrix}$ to both sides,

making the substitution $\alpha + b = \lambda$ and summing over $b$ using (<ref>),

renaming $\lambda$ as $\alpha$ and substituting the value of $c(n, j)$ from Proposition <ref>,
as before,  we get

\begin{eqnarray}
\label{Applied inductive steps in the hard inductive steps for g2-g1}
0 & \equiv & \sum_{\alpha = 0}^{p - 1}\sum_{j = r - n}^{\frac{r - 1}{2}}\frac{1}{p^{1.5}}\frac{(r - n)! n}{(r - j)\cdots (n - j)}{n - \frac{r - 3}{2} \choose n - r + j}(\alpha p)^{\frac{r + 3}{2} - j}(z - a -\alpha p)^j \mathbbm{1}_{a + \alpha p + p^2\Zp} \\
&  &  \qquad \qquad \mod \pi \latticeL{k} \nonumber
\end{eqnarray}

for each $\dfrac{r + 3}{2} \leq n \leq r$.

We wish to write the $a^{\mathrm{th}}$ summand on the right side of equation (<ref>) as the sum of $x_n \in E$ times (<ref>)  for  $\dfrac{r + 3}{2} \leq n \leq r$ and $x_{\frac{r+1}{2}} \in E$ times

\[
\sum_{\alpha = 0}^{p - 1}\frac{1}{p^{1.5}}(\alpha p)^{2}(z - a - \alpha p)^{\frac{r - 1}{2}}\mathbbm{1}_{a + \alpha p + p^2\Zp}.
\]

To solve for $x_{\frac{r+1}{2}}$, we compare the coefficients of $\sum\limits_{\alpha = 0}^{p - 1} \dfrac{1}{p^{1.5}}(\alpha p)^{\frac{r+3}{2}-j} (z - a - \alpha p)^{j}\mathbbm{1}_{a + \alpha p + p^2\Zp}$

for $0 \leq j \leq \dfrac{r-1}{2}$ to get the following matrix equation

\[
\! \begin{pmatrix}
{\frac{r + 3}{2} \choose 0}\frac{r}{r} & 0 & 0 & \cdots & 0 \\
{\frac{r + 3}{2} \choose 1}\frac{r}{r - 1} & {\frac{r + 1}{2} \choose 0}\frac{r - 1}{(r - 1)(r - 2)} & 0 & \cdots & 0 \\
{\frac{r + 3}{2} \choose 2}\frac{r}{r - 2} & {\frac{r + 1}{2} \choose 1}\frac{r - 1}{(r - 2)(r - 3)} & {\frac{r - 1}{2} \choose 0}\frac{2!(r - 2)}{(r - 2)(r - 3)(r - 4)} & \cdots & 0 \\
\vdots & \vdots & \vdots & & \vdots \\
\binom{\frac{r + 3}{2}}{\frac{r - 1}{2}}\frac{r}{\frac{r + 1}{2}} & {\frac{r + 1}{2} \choose \frac{r - 3}{2}}\frac{r - 1}{(\frac{r + 1}{2})(\frac{r - 1}{2})} & {\frac{r - 1}{2} \choose \frac{r - 5}{2}}\frac{2!(r - 2)}{(\frac{r + 1}{2})(\frac{r - 1}{2})(\frac{r - 3}{2})} & \cdots & 1
\end{pmatrix} \!\!
\begin{pmatrix}x_r \\ x_{r - 1} \\ x_{r - 2} \\ \vdots \\ x_{\frac{r+1}{2}} \end{pmatrix} \!\! = \!\! \begin{pmatrix}{\frac{r + 3}{2} \choose 0}\left(\cL - H_{\frac{r + 3}{2}}\right) \\ {\frac{r + 3}{2} \choose 1}\left(\cL - H_{\frac{r + 1}{2}}\right) \\ {\frac{r + 3}{2} \choose 2}\left(\cL - H_{\frac{r - 1}{2}}\right) \\ \vdots \\ {\frac{r + 3}{2} \choose \frac{r - 1}{2}}(\cL - H_2) \end{pmatrix}.
\]

It can be checked[
To save time and space we have  checked this for small
values of $r$ on a computer and extrapolated to all $r$.]  $x_{\frac{r+1}{2}} = (-1)^{\frac{r - 1}{2}}\dfrac{r + 3}{4}\left(\cL - H_{\frac{r - 1}{2}} - H_{\frac{r + 1}{2}}\right)$ (for $r > 1$). Therefore (<ref>) becomes

\begin{eqnarray*}
g_2(z) - g_1(z) \!\! & \!\! \equiv \!\! & \!\! \sum_{a = 0}^{p - 1}\sum_{\alpha = 0}^{p - 1}(-1)^{\frac{r - 1}{2}}\frac{r + 3}{4}\left(\cL - H_{\frac{r - 1}{2}} - H_{\frac{r + 1}{2}}\right)\frac{1}{p^{1.5}}(\alpha p)^2(z - a - \alpha p)^{\frac{r - 1}{2}}\mathbbm{1}_{a + \alpha p + p^2\Zp} \\
&& \qquad \qquad \mod \pi \latticeL{k}.
\end{eqnarray*}

Since $\nu > 0$, we see that the valuation of the coefficient of $(z - a - \alpha p)^{\frac{r - 1}{2}}\mathbbm{1}_{a + \alpha p + p^2\Zp}$ is strictly greater than $0.5$. Therefore Lemma <ref> implies that

\[
g_2(z) - g_1(z) \equiv 0 \mod \pi \latticeL{k}.
\]

Next, consider

\begin{eqnarray}\label{g1 in self-dual inductive step for n = r+3/2}
g_1(z) = \sum_{a = 0}^{p - 1}\left[\sum_{j = 0}^{\frac{r + 1}{2}}\frac{g^{(j)}(a)}{j!}(z - a)^j\right]\mathbbm{1}_{a + p\Zp}.
\end{eqnarray}

\[
\frac{g^{(j)}(a)}{j!} = {\frac{r + 3}{2} \choose j}\sum_{i = 0}^{p - 1}\frac{\lambda_i}{p^{1.5}}(a - i)^{\frac{r + 3}{2} - j}\logL(a - i).
\]

The $i = a$ term is $0$. For $i \neq a$, write $a - i = [a - i] + pa_i$ for some $a_i \in \Zp$. Note that $\logL(a - i) = \logL(1 + [a - i]^{-1}pa_i)$ since $[a - i]$ is a root of unity. Expanding $\logL(1 + [a - i]^{-1}pa_i)$, we get

\begin{eqnarray*}
\frac{g^{(j)}(a)}{j!} & \equiv & {\frac{r + 3}{2} \choose j}\sum_{\substack{i = 0  \\ i \neq a}}^{p - 1}\frac{\lambda_i}{p^{1.5}}[a - i]^{\frac{r + 1}{2} - j}pa_i \mod \pi \\
& \equiv & {\frac{r + 3}{2} \choose j}\sum_{\substack{i = 0 \\ i \neq a}}^{p - 1}\frac{\lambda_i}{p^{1.5}}\left(\frac{(a - i)^{\frac{r + 3}{2} - j} - [a - i]^{\frac{r + 3}{2} - j}}{\frac{r + 3}{2} - j}\right) \mod \pi.
\end{eqnarray*}

Recall that $\lambda_i = 1 \mod p$. Since the expression in the parentheses is divisible by $p$, we can replace $\lambda_i$ in the last line above by $1$. Moreover, using (<ref>), we get

\[
\frac{g^{(j)}(a)}{j!} \equiv {\frac{r + 3}{2} \choose j}\sum_{i = 0}^{p - 1}\frac{(a - i)^{\frac{r + 3}{2} - j}}{p^{1.5}\left(\frac{r + 3}{2} - j\right)} \mod \pi.
\]

Expanding $(a - i)^{\frac{r + 3}{2} - j} = \sum\limits_{l = 0}^{\frac{r + 3}{2} - j}{\frac{r + 3}{2} - j \choose l}(-1)^l i^la^{\frac{r + 3}{2} - j - l}$, we get

\[
\frac{g^{(j)}(a)}{j!} \equiv {\frac{r + 3}{2} \choose j}\sum_{l = 0}^{\frac{r + 3}{2} - j}\frac{(-1)^{l}}{\frac{r + 3}{2} - j}\frac{S_l}{p}{\frac{r + 3}{2} - j \choose l}\frac{a^{\frac{r + 3}{2} -j - l}}{p^{0.5}} \mod \pi,
\]

where $S_l$ is defined in equation (<ref>). Putting this equation in (<ref>), we get

\[
g_1(z) \equiv \sum_{a = 0}^{p - 1}\sum_{j = 0}^{\frac{r + 1}{2}}\sum_{l = 0}^{\frac{r + 3}{2} - j}\frac{(-1)^l}{\frac{r + 3}{2} - j} \frac{S_l}{p} {\frac{r + 3}{2} \choose j}{\frac{r + 3}{2} - j \choose l}\frac{a^{\frac{r + 3}{2} - j - l}}{p^{0.5}}(z - a)^j\mathbbm{1}_{a + p\Zp} \mod \pi \latticeL{k}.
\]

Since we have shown that $g(z) \equiv g_1(z) \mod \pi \latticeL{k}$, we get

\begin{eqnarray}\label{primitive hard inductive step for n = r+3/2}
\!\!\!\!\!\!\!\! 0 \equiv \sum_{a = 0}^{p - 1}\sum_{j = 0}^{\frac{r + 1}{2}}\sum_{l = 0}^{\frac{r + 3}{2} - j}\frac{(-1)^l}{\frac{r + 3}{2} - j} \frac{S_l}{p} {\frac{r + 3}{2} \choose j}{\frac{r + 3}{2} - j \choose l}\frac{a^{\frac{r + 3}{2} - j - l}}{p^{0.5}}(z - a)^j\mathbbm{1}_{a + p\Zp} \!\!\!\! \mod \pi \latticeL{k}.
\end{eqnarray}

Next, for $l = 0, 1, 2, \ldots, \dfrac{r + 3}{2}$, using the binomial expansion we have

\begin{eqnarray*}
\frac{1}{p^{0.5}}z^{\frac{r + 3}{2}-l}\mathbbm{1}_{\Zp} & \equiv & \sum_{a = 0}^{p - 1}\sum_{j = 0}^{\frac{r + 3}{2}-l}{\frac{r + 3}{2} -l \choose j} \frac{1}{p^{0.5}}a^{\frac{r + 3}{2}-l- j}(z - a)^j\mathbbm{1}_{a + p\Zp}
\end{eqnarray*}

\begin{eqnarray*}
\frac{1}{p^{0.5}}z^{\frac{r + 3}{2}}\mathbbm{1}_{\Zp} & \equiv & \sum_{a = 0}^{p - 1}\sum_{j = 0}^{\frac{r + 3}{2}}{\frac{r + 3}{2} \choose j} \frac{1}{p^{0.5}}a^{\frac{r + 3}{2}- j}(z - a)^j\mathbbm{1}_{a + p\Zp} \\
\frac{1}{p^{0.5}}z^{\frac{r + 1}{2}}\mathbbm{1}_{\Zp} & \equiv & \sum_{a = 0}^{p - 1}\sum_{j = 0}^{\frac{r + 1}{2}} {\frac{r + 1}{2} \choose j}\frac{1}{p^{0.5}}a^{\frac{r + 1}{2} - j}(z - a)^j\mathbbm{1}_{a + p\Zp} \\
\frac{1}{p^{0.5}}z^{\frac{r - 1}{2}}\mathbbm{1}_{\Zp} & \equiv & \sum_{a = 0}^{p - 1}\sum_{j = 0}^{\frac{r - 1}{2}}{\frac{r - 1}{2} \choose j}\frac{1}{p^{0.5}}a^{\frac{r - 1}{2} - j}(z - a)^{j}\mathbbm{1}_{a + p\Zp} \\
& \vdots & \\
\frac{1}{p^{0.5}}\mathbbm{1}_{\Zp} & \equiv & \sum_{a = 0}^{p - 1}\sum_{j = 0}^{0}{0 \choose j}\frac{1}{p^{0.5}}a^{-j}(z - a)^j\mathbbm{1}_{a + p\Zp}.
\end{eqnarray*}

Multiplying the $l^{\mathrm{th}}$ equation above by
$   \frac{(-1)^{l}}{\frac{r + 3}{2}}\frac{S_l}{p}{\frac{r + 3}{2} \choose l}, $
adding all the equations,

interchanging the sums over $l$ and $j$ and dropping the $j = \dfrac{r + 3}{2}$ term
using Lemma <ref>, we get

\begin{eqnarray*}
\sum_{l = 0}^{\frac{r + 3}{2}}\frac{(-1)^l}{\frac{r + 3}{2}}\frac{S_l}{p}{\frac{r + 3}{2} \choose l}\frac{1}{p^{0.5}}z^{\frac{r + 3}{2} - l}\mathbbm{1}_{\Zp} & \equiv & \sum_{a = 0}^{p - 1}\sum_{j = 0}^{\frac{r + 1}{2}}\sum_{l = 0}^{\frac{r + 3}{2} - j}\frac{(-1)^l}{\frac{r + 3}{2}}\frac{S_l}{p}{\frac{r + 3}{2} \choose l}{\frac{r + 3}{2} - l \choose j}\frac{a^{\frac{r + 3}{2} - l - j}}{p^{0.5}} \\
&& \qquad \qquad \qquad \qquad (z - a)^j\mathbbm{1}_{a + p\Zp} \mod \pi \latticeL{k}.
\end{eqnarray*}

Subtracting (<ref>) from the
equation above using

$    {\frac{r + 3}{2} \choose l}{\frac{r + 3}{2} - l \choose j} = {\frac{r + 3}{2} \choose j}{\frac{r + 3}{2} - j \choose l}$,

cancelling $\dfrac{r + 3}{2}$ on both sides, dropping the superfluous $j = 0$ summand on the
right side, and bringing the $1 \leq l \leq \dfrac{r + 3}{2}$ terms from the left side
to the right side, we get

\begin{eqnarray*}
\frac{1}{p^{0.5}}z^{\frac{r + 3}{2}}\mathbbm{1}_{\Zp} & \equiv & \sum_{l = 1}^{\frac{r + 3}{2}}(-1)^{l + 1}{\frac{r + 3}{2} \choose l}\frac{S_l}{p}\frac{1}{p^{0.5}}z^{\frac{r + 3}{2} - l}\mathbbm{1}_{\Zp} \\
&& + \sum_{a = 0}^{p - 1}\sum_{j = 1}^{\frac{r + 1}{2}}\sum_{l = 0}^{\frac{r + 3}{2} - j}(-1)^{l + 1}\frac{S_l}{p}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose j}{\frac{r + 3}{2} - j \choose l} \frac{a^{\frac{r + 3}{2} - j - l}}{p^{0.5}}(z - a)^j\mathbbm{1}_{a + p\Zp} \\
&& \qquad \qquad \mod \pi \latticeL{k}.
\end{eqnarray*}

Separating the $a = 0$ and $a \neq 0$ terms from the second sum above,
dropping the superfluous $a = 0$ terms

and applying $w$ to both sides,

we get

\begin{eqnarray*}
-\frac{1}{p^{0.5}}z^{\frac{r - 3}{2}}\mathbbm{1}_{p\Zp} & \equiv & \sum_{l = 1}^{\frac{r + 3}{2}}(-1)^{l}{\frac{r + 3}{2} \choose l}\frac{S_l}{p}\frac{1}{p^{0.5}}z^{\frac{r - 3}{2} + l}\mathbbm{1}_{p\Zp} \\
&& + \sum_{j = 1}^{\frac{r + 1}{2}} (-1)^{\frac{r + 3}{2} - j}\frac{S_{\frac{r + 3}{2} - j}}{p}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose \frac{r + 3}{2} - j}\frac{1}{p^{0.5}}z^{r - j}\mathbbm{1}_{\Zp} \\
&& + \sum_{a = 1}^{p - 1}\sum_{j = 1}^{\frac{r + 1}{2}}\sum_{l = 0}^{\frac{r + 3}{2} - j}(-1)^{l + 1}\frac{S_l}{p}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose j}{\frac{r + 3}{2} - j \choose l}\frac{a^{\frac{r + 3}{2} - j - l}}{p^{0.5}} \\
&& \qquad \qquad \qquad \qquad \qquad z^{r - j}(1 - az)^j\mathbbm{1}_{a^{-1} + p\Zp} \mod \pi \latticeL{k}.
\end{eqnarray*}

Splitting $\mathbbm{1}_{\Zp}$ in the second summand as a sum of $\mathbbm{1}_{p\Zp}$ and $\mathbbm{1}_{a^{-1} + p\Zp}$ for $1 \leq a \leq p - 1$, we get

\begin{eqnarray*}
-\frac{1}{p^{0.5}}z^{\frac{r - 3}{2}}\mathbbm{1}_{p\Zp} & \equiv & \sum_{l = 1}^{\frac{r + 3}{2}}(-1)^{l}{\frac{r + 3}{2} \choose l}\frac{S_l}{p}\frac{1}{p^{0.5}}z^{\frac{r - 3}{2} + l}\mathbbm{1}_{p\Zp} \\
&& + \sum_{j = 1}^{\frac{r + 1}{2}} (-1)^{\frac{r + 3}{2} - j}\frac{S_{\frac{r + 3}{2} - j}}{p}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose \frac{r + 3}{2} - j}\frac{1}{p^{0.5}}z^{r - j}\mathbbm{1}_{p\Zp} \\
&& + \sum_{a = 1}^{p - 1}\sum_{j = 1}^{\frac{r + 1}{2}}(-1)^{\frac{r + 3}{2} - j}\frac{S_{\frac{r + 3}{2} - j}}{p}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose \frac{r + 3}{2} - j}\frac{1}{p^{0.5}}z^{r - j}\mathbbm{1}_{a^{-1} + p\Zp} \\
&& + \sum_{a = 1}^{p - 1}\sum_{j = 1}^{\frac{r + 1}{2}}\sum_{l = 0}^{\frac{r + 3}{2} - j}(-1)^{l + 1}\frac{S_l}{p}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose j}{\frac{r + 3}{2} - j \choose l}\frac{a^{\frac{r + 3}{2} - j - l}}{p^{0.5}} \\
&& \quad \quad \quad \quad \quad \quad \quad \quad \qquad z^{r - j}(1 - az)^j\mathbbm{1}_{a^{-1} + p\Zp}\mod \pi \latticeL{k}.
\end{eqnarray*}

Using Lemma <ref>, we drop the $l > 2$ terms and the $j < \dfrac{r - 1}{2}$ terms from the first and the second sum on the right side, respectively, and combine them into the first sum below. Moreover, we expand $z^{r - j} = [(z - a^{-1}) + a^{-1}]^{r - j}$ in the third and the fourth sum and replace $a^{-1}$ by $b$ to get

\begin{eqnarray*}
-\frac{1}{p^{0.5}}z^{\frac{r - 3}{2}}\mathbbm{1}_{p\Zp} & \equiv & \sum_{l = 1}^{2}(-1)^l\frac{S_l}{p}{\frac{r + 3}{2} \choose l}\frac{r + 3}{2l}\frac{1}{p^{0.5}}z^{\frac{r - 3}{2} + l}\mathbbm{1}_{p\Zp} \\
&& + \sum_{b = 1}^{p - 1}\sum_{j = 1}^{\frac{r + 1}{2}}\sum_{m = 0}^{r - j}(-1)^{\frac{r + 3}{2} - j}\frac{S_{\frac{r + 3}{2} - j}}{p}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose \frac{r + 3}{2} - j}\frac{1}{p^{0.5}}{r - j \choose m}b^{m} \\
&& \quad \quad \quad \quad \quad \quad \>\> (z - b)^{r - j - m}\mathbbm{1}_{b + p\Zp} \\
&& + \sum_{b = 1}^{p - 1}\sum_{j = 1}^{\frac{r + 1}{2}}\sum_{l = 0}^{\frac{r + 3}{2} - j}\sum_{m = \frac{r - 1}{2}}^{r - j}(-1)^{l + j + 1}\frac{S_l}{p}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose j}{\frac{r + 3}{2} - j \choose l}{r - j \choose m} \\
&& \quad \quad \quad \quad \quad \quad \quad \quad \quad \qquad \frac{b^{l + m - \frac{r + 3}{2}}}{p^{0.5}}(z - b)^{r - m}\mathbbm{1}_{b + p\Zp} \mod \pi \latticeL{k}.
\end{eqnarray*}

We have dropped the $m < \dfrac{r - 1}{2}$ terms in the third line on the right side of the equation above using Lemma <ref>.
Applying the operator $\sum_{\alpha = 0}^{p - 1}\alpha^2\begin{pmatrix}1 & 0 \\ - a - \alpha p & p\end{pmatrix}$, we get

\begin{eqnarray}\label{Intermediate step in the hard inductive steps for n = r + 3/2}
& - & \sum_{\alpha = 0}^{p - 1}\alpha^2 p(z - a - \alpha p)^{\frac{r - 3}{2}}\mathbbm{1}_{a + \alpha p + p^2\Zp} \nonumber \\
& \equiv & \sum_{\alpha = 0}^{p - 1}\sum_{l = 1}^{2}\alpha^2p^{r/2}(-1)^l\frac{S_l}{p}{\frac{r + 3}{2}  \choose l}\frac{r + 3}{2l}\frac{1}{p^{0.5}}\left(\frac{z - a - \alpha p}{p}\right)^{\frac{r - 3}{2} + l}\mathbbm{1}_{a + \alpha p + p^2\Zp} \nonumber \\
&& + \sum_{\alpha = 0}^{p - 1}\sum_{b = 1}^{p - 1}\sum_{j = 1}^{\frac{r + 1}{2}}\sum_{m = 0}^{r - j}\alpha^2 p^{r/2}(-1)^{\frac{r + 3}{2} - j}\frac{S_{\frac{r + 3}{2} - j}}{p}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose \frac{r + 3}{2} - j}\frac{1}{p^{0.5}}{r - j \choose m}b^m \nonumber \\
&& \qquad\qquad\qquad\qquad\qquad \left(\frac{z - a - (\alpha + b)p}{p}\right)^{r - j - m}\mathbbm{1}_{a + (\alpha + b)p + p^2\Zp} \nonumber \\
&& + \sum_{\alpha = 0}^{p - 1}\sum_{b = 1}^{p - 1}\sum_{j = 1}^{\frac{r + 1}{2}}\sum_{l = 0}^{\frac{r + 3}{2} - j}\sum_{m = \frac{r - 1}{2}}^{r - j}\alpha^2 p^{r/2}(-1)^{l + j + 1}\frac{S_l}{p}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose j}{\frac{r + 3}{2} - j \choose l}{r - j \choose m}\frac{b^{l + m - \frac{r + 3}{2}}}{p^{0.5}} \nonumber \\
&& \qquad\qquad\qquad\qquad\qquad\qquad \left(\frac{z - a - (\alpha + b)p}{p}\right)^{r - m}\mathbbm{1}_{a + (\alpha + b)p + p^2\Zp} \!\!\mod \pi \latticeL{k}.
\end{eqnarray}

The first term on the right side of equation (<ref>) is the same in equation (<ref>). Next we evaluate the sums over $b$ in the remaining two sums on the right side of equation (<ref>) by substituting $\lambda = \alpha + b$ and using

\begin{eqnarray}\label{Sum over b identities for alpha^2}
\sum_{b = 1}^{p - 1}(\lambda - b)^2b^t =
\begin{cases}
0    & \text{ if }  1 - p < t < -2 \text{ or } 0 < t < p - 3\\
(-1)^{1-t}{2 \choose -t}\lambda^{2 + t}  & \text{ if } -2 \leq t \leq 0 \\
-1   & \text{ if } t = p - 3
\end{cases} \mod p.
\end{eqnarray}

Note that it is legal to do this by Lemma <ref>. This lemma
also allows us to replace $\lambda$ by its first $p$-adic digit.

First consider the summands in the second sum on the right for $0 \leq m \leq p - 3$. Note that the summands with $0 < m < p - 3$ are $0$ modulo $\pi\latticeL{k}$ by the first case in (<ref>). The third case in (<ref>) implies that the $m = p - 3$ summand is

\[
-\sum_{\lambda = 0}^{p - 1} \! p^{\frac{r - 1}{2}}\!(-1)^{\frac{r + 1}{2}}\!\frac{S_{\frac{r + 1}{2}}}{p}\frac{2}{r + 1}\frac{r + 3}{2}\mathbbm{1}_{a + \lambda p + p^2\Zp} \! \equiv \! -p^{\frac{r - 1}{2}}\!(-1)^{\frac{r + 1}{2}}\!\frac{S_{\frac{r + 1}{2}}}{p}\frac{r + 3}{r + 1}\mathbbm{1}_{a + p\Zp}\!\!\!\mod \pi \latticeL{k},
\]

which is $0$ modulo $\pi\latticeL{k}$ by Lemma <ref> noting $r > 1$.

Finally using the second case in (<ref>), we simplify the $m = 0$ summand and write it as the second line on the right side of equation (<ref>).
Next we analyze the summands in the third sum on the right. Note that the power of $b$ in this summand is in the range $[-2, p - 4]$. If $l > 2$, then it is in the range $[1, p - 4]$ so equation (<ref>) implies that these terms are zero modulo $\pi \latticeL{k}$. We consider the $l = 1, \> 2$ and $l = 0$ cases separately since retrospectively the $l = 0$ terms will be the important ones (cf. <ref>). First assume that $l = 1$ or $2$. The terms with the power of $b$ in the range $[1, p - 4]$ are $0$ modulo $\pi \latticeL{k}$ as seen above. The power of $b$ in the third sum on the right side of equation (<ref>) is $0$ only for the pairs $(l = 1, m = (r + 1)/2)$ and $(l = 2, m = (r - 1)/2)$ and is $-1$ only for $(l = 1, m = (r - 1)/2)$. The power of $b$ cannot be $-2$. Using equation (<ref>), we simplify these and write them as the third, fourth and fifth lines on the right below. Finally, using the first two cases of  (<ref>), we write the $l = 0$ summand of the third sum on the
right side of equation (<ref>) as the last two lines in equation (<ref>):

\begin{eqnarray}\label{six term identity for inductive steps for n = r+3/2 in self-dual}
& - & \sum_{\alpha = 0}^{p - 1}\alpha^2 p (z - a - \alpha p)^{\frac{r - 3}{2}}\mathbbm{1}_{a + \alpha p + p^2\Zp} \nonumber \\
& \equiv & \sum_{\alpha = 0}^{p - 1}\sum_{l = 1}^{2}\alpha^2p^{1 - l}(-1)^l\frac{S_l}{p}{\frac{r + 3}{2}  \choose l}\frac{r + 3}{2l}\left(z - a - \alpha p\right)^{\frac{r - 3}{2} + l}\mathbbm{1}_{a + \alpha p + p^2\Zp} \nonumber \\
&& - \sum_{\lambda = 0}^{p - 1}\sum_{j = 1}^{\frac{r + 1}{2}}p^{j - \frac{r + 1}{2}}(-1)^{\frac{r + 3}{2} - j}\frac{S_{\frac{r + 3}{2} - j}}{p}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose \frac{r + 3}{2} - j}\lambda^2(z - a - \lambda p)^{r - j}\mathbbm{1}_{a + \lambda p + p^2\Zp} \nonumber \\
&& - \sum_{\lambda = 0}^{p - 1}\sum_{j = 1}^{\frac{r - 1}{2}}\lambda^2 (-1)^j\frac{S_1}{p}j{\frac{r + 3}{2} \choose j}{r - j \choose \frac{r + 1}{2}}(z - a - \lambda p)^{\frac{r - 1}{2}}\mathbbm{1}_{a + \lambda p + p^2\Zp} \nonumber \\
&& - \sum_{\lambda = 0}^{p - 1}\sum_{j = 1}^{\frac{r - 1}{2}}\lambda^2 p^{-1} (-1)^{j + 1}\frac{S_2}{p}\frac{j\left(\frac{r + 1}{2} - j\right)}{2}{\frac{r + 3}{2} \choose j}{r - j \choose \frac{r - 1}{2}}(z - a - \lambda p)^{\frac{r + 1}{2}}\mathbbm{1}_{a + \lambda p + p^2\Zp} \nonumber \\
&& + \sum_{\lambda = 0}^{p - 1}\sum_{j = 1}^{\frac{r + 1}{2}}2\lambda p^{-1}(-1)^j \frac{S_1}{p}j{\frac{r + 3}{2} \choose j}{r - j \choose \frac{r - 1}{2}}(z - a - \lambda p)^{\frac{r + 1}{2}}\mathbbm{1}_{a + \lambda p + p^2\Zp} \nonumber \\
&& + \sum_{\lambda = 0}^{p - 1}\sum_{j = 1}^{\frac{r + 1}{2}}\sum_{m = \frac{r - 1}{2}}^{r - j}{2 \choose \frac{r + 3}{2} - m}\lambda^{m - \frac{r - 1}{2}}p^{m - \frac{r + 1}{2}}(-1)^{j + \frac{r + 3}{2} - m}\frac{S_0}{p}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose j}{r - j \choose m} \nonumber \\
&& \qquad \qquad \qquad \qquad\qquad (z - a - \lambda p)^{r - m}\mathbbm{1}_{a + \lambda p + p^2\Zp} \mod \pi \latticeL{k}.
\end{eqnarray}

We replace $\lambda$ by $\alpha$ in the last five sums on the right side of the equation above. We now check that the first four sums on the right side of (<ref>) are $0$ modulo $\pi \latticeL{k}$. First, using Lemma <ref> we see that the terms with $j < \dfrac{r - 1}{2}$ in the second sum on the right side of (<ref>) are $0$ modulo $\pi\latticeL{k}$. Therefore the first four terms above only involve the $\dfrac{r - 1}{2}$ and $\dfrac{r + 1}{2}$ powers of $(z - a - \alpha p)$. Using equation (<ref>), we see that the sum of the coefficients of $\alpha^2(z - a - \alpha p)^{\frac{r - 1}{2}}\mathbbm{1}_{a + \alpha p + p^2\Zp}$ in these four terms is

\begin{eqnarray*}
&& \frac{1}{2}\left(\frac{r + 3}{2}\right)^2 - \frac{1}{2}\left(\frac{r + 1}{2}\right)\left(\frac{r + 3}{2}\right) + \sum_{j = 1}^{\frac{r - 1}{2}}(-1)^j\left(\frac{1}{2}\right)j{\frac{r + 3}{2} \choose j}{r - j \choose \frac{r + 1}{2}} \\
&& = \frac{1}{2}\left(\frac{r + 3}{2}\right)\left[1 + \sum_{j = 1}^{\frac{r - 1}{2}}(-1)^j{\frac{r + 1}{2} \choose j - 1}{r - j \choose \frac{r + 1}{2}}\right] = 0,
\end{eqnarray*}

by <cit.> with $k = \dfrac{r - 3}{2}$, $n = \dfrac{r + 1}{2}$ and $i = j - 1$ there.
Next, using equation (<ref>), the sum of coefficients of $\alpha^2 p^{-1}(z - a - \alpha p)^{\frac{r + 1}{2}}\mathbbm{1}_{a + \alpha p + p^2\Zp}$ in the same four terms is

\begin{eqnarray*}
&& \left(\frac{1}{6}\right)\left(\frac{(r + 3)(r + 1)}{8}\right)\left(\frac{r + 3}{4}\right) - \left(\frac{1}{6}\right)\left(\frac{r - 1}{4}\right)\left(\frac{(r + 3)(r + 1)}{8}\right)  \\
&& + \sum_{j = 1}^{\frac{r - 1}{2}}(-1)^j\left(\frac{1}{6}\right)\frac{j\left(\frac{r + 1}{2} - j\right)}{2}{\frac{r + 3}{2} \choose j}{r - j \choose \frac{r - 1}{2}} \\
&& = \left(\frac{1}{6}\right)\left(\frac{(r + 3)(r + 1)}{8}\right) \left[1 + \sum_{j = 1}^{\frac{r - 1}{2}}(-1)^j\frac{\frac{r + 1}{2} - j}{\frac{r + 3}{2} - j}{\frac{r - 1}{2} \choose j - 1}{r - j \choose \frac{r - 1}{2}}\right] = 0.
\end{eqnarray*}

Indeed, consider

\begin{eqnarray*}
&  & \sum_{j = 1}^{\frac{r - 1}{2}}(-1)^j\frac{\frac{r + 1}{2} - j}{\frac{r + 3}{2} - j}{\frac{r - 1}{2} \choose j - 1}{r - j \choose \frac{r - 1}{2}} \\
%&  &  =  \sum_{j = 1}^{\frac{r - 1}{2}}(-1)^j{\frac{r - 1}{2} \choose j - 1}{r - j \choose \frac{r - 1}{2}} - \sum_{j = 1}^{\frac{r - 1}{2}}(-1)^{j}\frac{1}{\frac{r + 3}{2} - j}{\frac{r - 1}{2} \choose \frac{r + 1}{2} - j}{r - j \choose \frac{r - 1}{2}} \\
&  & = \sum_{j = 1}^{\frac{r - 1}{2}}(-1)^j{\frac{r - 1}{2} \choose j - 1}{r - j \choose \frac{r - 1}{2}} - \frac{1}{\frac{r + 1}{2}}\sum_{j = 1}^{\frac{r - 1}{2}}(-1)^j{\frac{r + 1}{2} \choose j - 1}{r - j \choose \frac{r - 1}{2}}.
\end{eqnarray*}

Putting $k = \dfrac{r - 1}{2}, \> n = \dfrac{r - 1}{2}$, and $i = j - 1$ in <cit.>, the first sum on the right side of the equation above is $(-1)^{\frac{r - 1}{2}} - 1$. Putting $k = j - 1, \> n = \dfrac{r - 1}{2}, \> x = -1$ and $y = \dfrac{r - 1}{2}$ in <cit.>, the second sum is $-(-1)^{\frac{r + 1}{2}}\left(\dfrac{r + 1}{2}\right)$.

Next, the fifth term on the right side of equation (<ref>) vanishes because

\[
\sum_{j = 1}^{\frac{r + 1}{2}}(-1)^jj{\frac{r + 3}{2} \choose j}{r - j \choose \frac{r - 1}{2}} = \left(\frac{r + 3}{2}\right)\sum_{j = 1}^{\frac{r + 1}{2}}(-1)^j{\frac{r + 1}{2} \choose j - 1}{r - j \choose \frac{r - 1}{2}} = 0,
\]

by <cit.> with $k = j - 1$, $n = \dfrac{r - 1}{2}$, $x = -1$ and $y = \dfrac{r - 1}{2}$.

Therefore equation (<ref>) becomes

\begin{eqnarray}\label{l = 0 are the important summands}
&&-\sum_{\alpha = 0}^{p - 1}\alpha^2 p(z - a - \alpha p)^{\frac{r - 3}{2}}\mathbbm{1}_{a + \alpha p + p^2\Zp} \\
&& \equiv \sum_{\alpha = 0}^{p - 1}\sum_{j = 1}^{\frac{r + 1}{2}}\sum_{m = \frac{r - 1}{2}}^{r - j}{2 \choose \frac{r + 3}{2} - m}\alpha^{m - \frac{r - 1}{2}}p^{m - \frac{r + 1}{2}}(-1)^{j + \frac{r + 3}{2} - m}\frac{S_0}{p}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose j}{r - j \choose m} \nonumber \\
&& \qquad \qquad \qquad \qquad \qquad (z - a - \alpha p)^{r - m}\mathbbm{1}_{a + \alpha p + p^2\Zp} \mod \pi \latticeL{k}. \nonumber
\end{eqnarray}

Interchanging the $j$ and $m$ sums and replacing $m$ by $r - l$, we get

\begin{eqnarray*}
&&-\sum_{\alpha = 0}^{p - 1}\alpha^2 p(z - a - \alpha p)^{\frac{r - 3}{2}}\mathbbm{1}_{a + \alpha p + p^2\Zp} \\
% && \equiv \sum_{\alpha = 0}^{p - 1}\sum_{l = 1}^{\frac{r + 1}{2}}\sum_{j = 1}^{l}{2 \choose l - \frac{r - 3}{2}}\alpha^{\frac{r + 1}{2} - l}p^{\frac{r - 1}{2} - l}(-1)^{j - \frac{r - 3}{2} + l}\frac{S_0}{p}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose j}{r - j \choose r - l} \nonumber \\
% && \qquad \qquad \qquad \qquad(z - a - \alpha p)^{l}\mathbbm{1}_{a + \alpha p + p^2\Zp} \\
&& \equiv \sum_{\alpha = 0}^{p - 1}\sum_{l = 1}^{\frac{r + 1}{2}}{2 \choose l - \frac{r - 3}{2}}\alpha^{\frac{r + 1}{2} - l}p^{\frac{r - 1}{2} - l}(-1)^{l - \frac{r - 3}{2}}\left(\sum_{j = 1}^{l}(-1)^{j}\frac{j}{\frac{r + 3}{2} - j}{\frac{r + 3}{2} \choose j}{r - j \choose r - l}\right) \\
&& \qquad\qquad\qquad\qquad (z - a -\alpha p)^{l}\mathbbm{1}_{a + \alpha p + p^2\Zp} \mod \latticeL{k}.
\end{eqnarray*}

The expression inside the parentheses is $c(n, l)$
appearing in the proof of Proposition <ref>  with $n = \dfrac{r + 3}{2}$.
Substitute its value obtained in that proposition. The $-1$ part of $c(n, l)$ with
$l = r - n$ cancels with the term on the left side. Therefore we obtain the proposition for $n = \dfrac{r + 3}{2}$.

The following lemma is used in the proof of Proposition <ref>.

Let $1 \leq r \leq p-2$ be odd.
For $\nu > -0.5$, we have

\[
\sum_{a = 0}^{p - 1}(z - a)^{\frac{r - 1}{2}}\mathbbm{1}_{a + p\Zp} \equiv 0 \mod \pi \latticeL{k}.
\]

The statement follows from Lemma <ref> for $r = 1$.
If $3 \leq r \leq p-2$, the proof is similar to the proof of Proposition <ref>
and is left as an exercise (or see the first author's thesis): one uses

\[
g(z) = p^{-1}\sum_{i \in I} \lambda_i (z - i)^{\frac{r + 3}{2}}\logL(z - i),
\]

where $I = \left\{0, \> 1, \> \ldots, \> \dfrac{r + 3}{2}, \> p\right\}$ and $\lambda_i \in \zp$
are as in Lemma <ref> (2).

\begin{eqnarray}\label{Sum identity in killing T12 in self-dual}
\sum_{i \in I}\lambda_i i^j = 0 \text{ for } 0 \leq j \leq \frac{r + 3}{2}
\end{eqnarray}

\begin{eqnarray}\label{Congruence identity for coefficients in killing T12 in self-dual}
\lambda_0 \equiv 1 \!\!\!\!\mod p, \>  \lambda_i \equiv 0 \!\!\!\!\mod p \text{ for } 1 \leq i \leq \frac{r + 3}{2} \text{ and } \lambda_{p} = - 1.
\end{eqnarray}

Write $g(z) = g(z)\mathbbm{1}_{\Zp} + g(z)\mathbbm{1}_{\Qp \setminus \Zp}$. The usual computation shows that

\[
w \cdot g(z)\mathbbm{1}_{\Qp \setminus \Zp} = \sum_{i \in I}p^{-1}\lambda_iz^{\frac{r - 3}{2}}(1 - zi)^{\frac{r + 3}{2}}\logL(1 - zi)\mathbbm{1}_{p\Zp}.
\]

Lemma <ref> implies that $g(z)\mathbbm{1}_{\Qp \setminus \Zp} \equiv 0 \mod \pi \latticeL{k}$. So $g(z) \equiv g(z)\mathbbm{1}_{\Zp} \mod \pi \latticeL{k}$.

Using Lemma <ref> (1), we see that $g(z)\mathbbm{1}_{\Zp} \equiv g_2(z) \mod \pi \latticeL{k}$. We analyze $g_2(z) - g_1(z)$ and $g_1(z)$ separately.

Recall that for $0 \leq j \leq \dfrac{r + 1}{2}$, the coefficient of $(z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2\Zp}$ in $g_2(z) - g_1(z)$ is

\begin{eqnarray}\label{jth summand in g2 - g1 in killing T12 in self-dual}
&& {\frac{r + 3}{2} \choose j}\sum_{i \in I}p^{-1}\lambda_i\Big[(a + \alpha p - i)^{\frac{r + 3}{2} - j}\logL(a + \alpha p - i) - (a - i)^{\frac{r + 3}{2} - j}\logL(a - i) \\
&& - {\frac{r + 3}{2} - j \choose 1}(\alpha p)(a - i)^{\frac{r + 1}{2} - j}\logL(a - i) - \cdots - {\frac{r + 3}{2} - j \choose \frac{r + 1}{2} - j}(\alpha p)^{\frac{r + 1}{2} - j}(a - i)\logL(a -  i)\Big]. \nonumber
\end{eqnarray}

We prove that this coefficient is congruent to $0$ modulo $p^{r/2 - j}\pi$.

First assume that $a \neq 0$. For $i \neq a$, the $i^{\mathrm{th}}$ summand in expression (<ref>) can be written as

\[
p^{-1}\lambda_i\big[(a - i + \alpha p)^{\frac{r + 3}{2} - j}\logL(1 + (a - i)^{-1}\alpha p) + (\alpha p)^{\frac{r + 3}{2} - j}\logL(a - i)\big].
\]

The conditions $i \neq a$ and $1 \leq a \leq p - 1$ imply that $a - i$ is a $p$-adic unit. Therefore the valuation of the last term in the expression above is greater than $r/2 - j$. So we drop the last term modulo $p^{r/2 - j}\pi$. Moreover, expanding $(a - i + \alpha p)^{\frac{r + 3}{2} - j}$ and the logarithm, the expression above becomes

\begin{eqnarray}\label{a neq 0 and zi neq a term in killing T12 in self-dual}
&& p^{-1}\lambda_i\big[(a - i)^{\frac{r + 1}{2} - j}\alpha p + c_{\frac{r - 1}{2} - j}(a - i)^{\frac{r - 1}{2} - j}(\alpha p)^2 + \cdots + c_1(a - i)(\alpha p)^{\frac{r + 1}{2} - j}\big] \\
&& \qquad \qquad \mod p^{r/2 - j}\pi, \nonumber
\end{eqnarray}

for some $c_1,  \ldots,  c_{\frac{r - 1}{2} - j} \in \Zp$.
Next assume that $i = a$. Then the $i^{\mathrm{th}}$ summand in expression (<ref>) can be written as

\[
p^{-1}\lambda_i(\alpha p)^{\frac{r + 3}{2} - j}\logL(\alpha p).
\]

Since $v_p(\cL) > -0.5$, the valuation of this term is greater than $r/2 - j$. Therefore the $i = a$ term in expression (<ref>) is congruent to $0$ modulo $p^{r/2 - j}\pi$.

Plugging expression (<ref>) in to expression (<ref>), we see that the coefficient of $(z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2 \Zp}$ in $g_2(z) - g_1(z)$ is $0$ modulo $p^{r/2 - j}\pi$ if $a \neq 0$.

Finally assume that $a = 0$. Using $v_p(\cL) > -0.5$, we see that each term in expression (<ref>) for $i = 0$ and $p$ has valuation greater than $r/2 - j$. Therefore for $a = 0$, the $i = 0$ and $p$ summands in expression (<ref>) are $0$ modulo $p^{r/2 - j}\pi$. For $i \neq 0, p$, the $i^{\mathrm{th}}$ summand in expression (<ref>) is also given by (<ref>).
Using (<ref>), the coefficient of $(z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2 \Zp}$ in $g_2(z) - g_1(z)$ is

\[
-{\frac{r + 3}{2} \choose j}p^{-1}\lambda_{p}\left[(-p)^{\frac{r + 1}{2} - j}\alpha p + t_{\frac{r - 1}{2} - j}(-p)^{\frac{r - 1}{2} - j}(\alpha p)^2 + \cdots + t_1(-p)(\alpha p)^{\frac{r + 1}{2} - j}\right] \!\! \mod p^{r/2 - j}\pi.
\]

Each term in the expression above has valuation greater than $r/2 - j$. Therefore we have proved that the coefficient of $(z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2\Zp}$ in $g_2(z) - g_1(z)$ is $0$ modulo $p^{r/2 - j}$ if $a = 0$.

Next consider

\begin{eqnarray}\label{g1 in killing T12 in self-dual}
g_1(z) = \sum_{a = 0}^{p - 1} \left[\sum_{j = 0}^{\frac{r + 1}{2}}\frac{g^{(j)}(a)}{j!}(z - a)^j \right] \mathbbm{1}_{a + p\Zp},
\end{eqnarray}

\begin{eqnarray}\label{jth summand in g1 in killing T12 in self-dual}
\frac{g^{(j)}(a)}{j!} = {\frac{r + 3}{2} \choose j}\sum_{i \in I}p^{-1}\lambda_i(a - i)^{\frac{r + 3}{2} - j}\logL(a - i).
\end{eqnarray}

First assume that $a \neq 0$. The $i \neq 0, a, p$ terms on the right side of equation (<ref>) are congruent to $0$ modulo $\pi$ since $a \not \equiv i \!\! \mod p$ and $p \mid \lambda_i$ for $i \neq 0, p$ by (<ref>). The $i = a$ term is equal to $0$. The sum of the $i = 0$ and $i = p$ summands on the right side of (<ref>) is

\[
p^{-1} \left[\lambda_0a^{\frac{r + 3}{2} - j}\logL(a) + \lambda_{p}(a - p)^{\frac{r + 3}{2} - j}\logL(a - p)\right].
\]

Recall from (<ref>) that $\lambda_0 = 1 \!\! \mod p$ and $\lambda_{p} = -1$. Therefore the display above is congruent to

\[
p^{-1}\left[a^{\frac{r + 3}{2} - j}\logL(a) - (a - p)^{\frac{r + 3}{2} - j}\logL(a - p)\right] \mod \pi.
\]

We expand $(a - p)^{\frac{r + 3}{2} - j}$ and drop the terms divisible by $p$ since $p \> \vert \> \logL(a - p)$ to get

\[
p^{-1}(-a^{\frac{r + 3}{2} - j})\logL(1 - a^{-1}p) \mod \pi.
\]

Expanding $\logL(1 - a^{-1}p)$
and dropping the terms that are congruent to $0$ mod $\pi$, we see that the sum of the $i = 0, \> p$ summands on the right side of (<ref>) is

\[
a^{\frac{r + 1}{2} - j} \mod \pi.
\]

Therefore for $a \neq 0$, we get

\begin{eqnarray}\label{jth term in a neq 0 in g1 in killing T12 in self-dual}
\frac{g^{(j)}(a)}{j!} \equiv {\frac{r + 3}{2} \choose j}a^{\frac{r + 1}{2} - j} \mod \pi.
\end{eqnarray}

Next, assume that $a = 0$. Then again the $i \neq 0, p$ terms on the right side of equation (<ref>) are congruent to $0$ modulo $\pi$ by the same reasoning as in the $a \neq 0$ case above. The $i = 0$ term is $0$. The $i = p$ summand
on the right side of equation (<ref>) is

\[
-p^{-1}(-p)^{\frac{r + 3}{2} - j}\cL.
\]

Therefore for $a = 0$, we have

\begin{eqnarray}\label{jth term in a = 0 in g1 in killing T12 in self-dual}
\frac{g^{(j)}(0)}{j!} \equiv (-1)^{\frac{r + 5}{2} - j}{\frac{r + 3}{2} \choose j}p^{\frac{r + 1}{2} - j}\cL \mod \pi.
\end{eqnarray}

Putting equations (<ref>) and (<ref>) in equation (<ref>) and using Lemma <ref>, we get

\begin{eqnarray*}
g_1(z) & \equiv & \sum_{a = 1}^{p - 1}\left[\sum_{j = 0}^{\frac{r + 1}{2}}{\frac{r + 3}{2} \choose j}a^{\frac{r + 1}{2} - j}(z - a)^j\right]\mathbbm{1}_{a + p\Zp} + \left[\sum_{j = 0}^{\frac{r + 1}{2}}(-1)^{\frac{r + 5}{2} - j}{\frac{r + 3}{2} \choose j}p^{\frac{r + 1}{2} - j}\cL z^j\right]\mathbbm{1}_{p\Zp} \\
& \equiv & \sum_{a = 1}^{p - 1}a^{-1}\left[z^{\frac{r + 3}{2}} - (z - a)^{\frac{r + 3}{2}}\right]\mathbbm{1}_{a + p\Zp} + \left[\sum_{j = 0}^{\frac{r + 1}{2}}(-1)^{\frac{r + 5}{2} - j}{\frac{r + 3}{2} \choose j}p^{\frac{r + 1}{2} - j}\cL z^j\right]\mathbbm{1}_{p\Zp} \\
&& \qquad \qquad \mod \pi \latticeL{k}.
\end{eqnarray*}

Using Lemma <ref> to kill
the second term in the first sum and the $j = \dfrac{r + 1}{2}$ summand in the second sum,
and Lemma <ref> to kill the other terms in the second
sum, mod $\pi \latticeL{k}$, we get

\[
g_1(z) \equiv \sum_{a = 1}^{p - 1}a^{-1}z^{\frac{r + 3}{2}}\mathbbm{1}_{a + p\Zp} \mod \pi \latticeL{k}.
\]

Since $g(z) \equiv g_1(z) \mod \pi \latticeL{k}$, we get

\[
0 \equiv \sum_{a = 1}^{p - 1}a^{-1}z^{\frac{r + 3}{2}}\mathbbm{1}_{a + p\Zp} \mod \pi \latticeL{k}.
\]

Applying $\beta$,
we get

\[
0 \equiv \sum_{a = 1}^{p - 1} a^{-1}p^{3/2}z^{\frac{r - 3}{2}}\mathbbm{1}_{a^{-1}p + p^2\Zp} \mod \pi \latticeL{k}.
\]

Expanding $z^{\frac{r - 3}{2}} = \sum\limits_{j = 0}^{\frac{r - 3}{2}}{\frac{r - 3}{2} \choose j}(a^{-1}p)^{\frac{r - 3}{2} - j}(z - a^{-1}p)^j$, replacing $a^{-1}$ by $a$, and
including  the superfluous $a = 0$ term, we get

\begin{eqnarray}\label{Main congruence in killing T12 in self-dual}
0 \equiv \sum_{a = 0}^{p - 1}\sum_{j = 0}^{\frac{r - 3}{2}}{\frac{r - 3}{2} \choose j}p^{r/2 - j}a^{\frac{r - 1}{2} - j}(z - ap)^j\mathbbm{1}_{ap + p^2\Zp} \mod \pi \latticeL{k}.
\end{eqnarray}

Next, the shallow inductive steps given in Proposition <ref> are

\[
z^{r - n}\mathbbm{1}_{p\Zp} \equiv \sum_{b = 1}^{p - 1}\sum_{j = 1}^{\frac{r - 1}{2}}c(n, j)b^{r - j - n}(z - b)^j\mathbbm{1}_{b + p\Zp} \mod \pi \latticeL{k},
\]

for $\dfrac{r + 3}{2} \leq n \leq r$. Applying $\sum\limits_{a = 0}^{p - 1}a^{n - \frac{r + 1}{2}}\begin{pmatrix}1 & 0 \\ -pa & p\end{pmatrix}$, we get

\begin{eqnarray*}
\sum_{a = 0}^{p - 1}a^{n - \frac{r + 1}{2}} p^{n - r/2}(z - ap)^{r - n}\mathbbm{1}_{ap + p^2\Zp} & \equiv & \sum_{a = 0}^{p - 1}\sum_{b = 1}^{p - 1}\sum_{j = 1}^{\frac{r - 1}{2}}p^{r/2 - j}c(n, j)a^{n - \frac{r + 1}{2}}b^{r - j - n} \\
&& \quad \quad \quad (z - (a + b)p)^j\mathbbm{1}_{(a + b)p + p^2\Zp} \mod \pi \latticeL{k}.
\end{eqnarray*}

Substituting $\lambda = a + b$ and evaluating the sum over $b$ using the identity (<ref>) with $i = \dfrac{r + 3}{2}$, we get

\begin{eqnarray*}
\sum_{a = 0}^{p - 1} a^{n - \frac{r + 1}{2}} p^{n - r/2}(z - ap)^{r - n}\mathbbm{1}_{ap + p^2\Zp} & \equiv & \sum_{\lambda = 0}^{p - 1}\sum_{j = r - n}^{\frac{r - 1}{2}}(-1)^{n - r + j + 1}p^{r/2 - j}{n - \frac{r + 1}{2} \choose n - r + j}c(n, j)\lambda^{\frac{r - 1}{2} - j} \\
&& \quad \quad \quad \quad \quad (z - \lambda p)^j\mathbbm{1}_{\lambda p + p^2\Zp} \mod \pi \latticeL{k}.
\end{eqnarray*}

Replacing $\lambda$ by $a$ and substituting $c(n, j)$ from Proposition <ref>,

we get

\begin{eqnarray}\label{Refined inductive steps in killing T12 in self-dual}
\! 0 \! \equiv \! \sum_{a = 0}^{p - 1}\sum_{j = r - n}^{\frac{r - 1}{2}}\!\!\!{n - \frac{r + 1}{2} \choose n - r + j}\!\frac{(r - n)! n}{(r - j)\cdots (n - j)}p^{r/2 - j}a^{\frac{r - 1}{2} - j}(z - ap)^j\mathbbm{1}_{ap + p^2\Zp} \!\!\! \mod \pi \latticeL{k}.
\end{eqnarray}

We wish to write $p^{0.5}(z - ap)^{\frac{r - 1}{2}}\mathbbm{1}_{ap + p^2\Zp}$ as a sum of $x_n \in E$ times the $a^{\mathrm{th}}$ summand on the right side of (<ref>) for $\dfrac{r + 3}{2} \leq n \leq r$, and $x_{\frac{r + 1}{2}} \in E$ times the corresponding summand in (<ref>). Comparing coefficients of $p^{r/2 - j}a^{\frac{r - 1}{2} - j}(z - ap)^{j}\mathbbm{1}_{ap + p^2\Zp}$, we get the following matrix equation

\begin{eqnarray}
\begin{pmatrix}
{\frac{r - 1}{2} \choose 0}\frac{r}{r} & 0 & 0 & \cdots & {\frac{r - 3}{2} \choose 0} \\
{\frac{r - 1}{2} \choose 1}\frac{r}{r - 1} & {\frac{r - 3}{2} \choose 0}\frac{r - 1}{(r - 1)(r - 2)} & 0 & \cdots & {\frac{r - 3}{2} \choose 1} \\
{\frac{r - 1}{2} \choose 2}\frac{r}{r - 2} & {\frac{r - 3}{2} \choose 1}\frac{r - 1}{(r - 2)(r - 3)} & {\frac{r - 5}{2} \choose 0}\frac{2!(r - 2)}{(r - 2)(r - 3)(r - 4)} & \cdots & {\frac{r - 3}{2} \choose 2} \\
\vdots & \vdots & \vdots & & \vdots \\
{\frac{r - 1}{2} \choose \frac{r - 3}{2}}\frac{r}{\frac{r + 3}{2}} & {\frac{r - 3}{2} \choose \frac{r - 5}{2}}\frac{r - 1}{(\frac{r + 3}{2})(\frac{r + 1}{2})} & {\frac{r - 5}{2} \choose \frac{r - 7}{2}}\frac{2!(r - 2)}{(\frac{r + 3}{2})(\frac{r + 1}{2})(\frac{r - 1}{2})} & \cdots & {\frac{r - 3}{2} \choose \frac{r - 3}{2}} \\
{\frac{r - 1}{2} \choose \frac{r - 1}{2}}\frac{r}{\frac{r + 1}{2}} & {\frac{r - 3}{2} \choose \frac{r - 3}{2}}\frac{r - 1}{(\frac{r + 1}{2})(\frac{r - 1}{2})} & {\frac{r - 5}{2} \choose \frac{r - 5}{2}}\frac{2!(r - 2)}{(\frac{r + 1}{2})(\frac{r - 1}{2})(\frac{r - 3}{2})} & \cdots & 0
\end{pmatrix}
\begin{pmatrix}
x_r \\ x_{r - 1} \\ x_{r - 2} \\ \vdots \\ x_{\frac{r + 1}{2}}
\end{pmatrix}
= \begin{pmatrix}
0 \\ 0 \\ 0 \\ \vdots \\ 0 \\ 1
\end{pmatrix}.
\end{eqnarray}

We claim that in the solution to this matrix equation, $x_n \in \Zp$ for all $\dfrac{r + 1}{2} \leq n \leq r$. To show this, it is enough to prove that $\det A \in \Zp^*$. Note that $A$ is almost lower triangular with all but the last diagonal entry belonging to $\Zp^*$. We show that for $0 \leq j \leq \dfrac{r - 3}{2}$ after making the $\left(j, \dfrac{r - 1}{2}\right)^{\mathrm{th}}$ entries $0$ using column operations, the $\left(\dfrac{r - 1}{2}, \dfrac{r - 1}{2}\right)^{\mathrm{th}}$ entry is a $p$-adic unit.

First perform the operation $C_{\frac{r - 1}{2}} \to C_{\frac{r - 1}{2}} - C_0$ to make the $\left(0, \dfrac{r - 1}{2}\right)^{\mathrm{th}}$ entry in the last column $0$. Therefore for $j \geq 1$, the $\left(j, \dfrac{r - 1}{2}\right)^{\mathrm{th}}$ entry becomes

\[
{\frac{r - 3}{2} \choose j} - {\frac{r - 1}{2} \choose j}\frac{r}{r - j} = -{\frac{r - 3}{2} \choose j - 1}\frac{r + \frac{r - 1}{2} - j}{r - j}.
\]

Next, perform the operation $C_{\frac{r - 1}{2}} \to C_{\frac{r - 1}{2}} + \frac{(r - 2)\left(r + \frac{r - 1}{2} - 1\right)}{r - 1}C_1$ to make the $\left(1, \dfrac{r - 1}{2}\right)^{\mathrm{th}}$ entry in the last column $0$. Therefore for $j \geq 2$, the $\left(j, \dfrac{r - 1}{2}\right)^{\mathrm{th}}$ entry becomes

\begin{eqnarray*}
&&-{\frac{r - 3}{2} \choose j - 1}\frac{r + \frac{r - 1}{2} - j}{r - j} + \frac{(r - 2)\left(r + \frac{r - 1}{2} - 1\right)}{r - 1}{\frac{r - 3}{2} \choose j - 1}\frac{r - 1}{(r - j)(r - j - 1)} \\
&& = {\frac{r - 5}{2} \choose j - 2}\left(\frac{r - 3}{2}\right)\frac{2(r - 1) + \frac{r - 1}{2} - j}{(r - j)(r - j - 1)}.
\end{eqnarray*}

Continuing this process, after making the $\left(\dfrac{r - 3}{2}, \dfrac{r - 1}{2}\right)^{\mathrm{th}}$ entry $0$, we see that for $j \geq \dfrac{r - 1}{2}$ the $\left(j, \dfrac{r - 1}{2}\right)^{\mathrm{th}}$ entry in the last column is

\[
(-1)^{\frac{r - 1}{2}}{0 \choose j - \frac{r - 1}{2}}\left(\frac{r - 3}{2}\right)!\frac{\left(\frac{r - 1}{2}\right)\left(r - \frac{r - 3}{2}\right) + \left(\frac{r - 1}{2} - j\right)}{(r - j)(r - j - 1) \cdots \left(r - j - \frac{r - 3}{2}\right)}.
\]

Therefore the $\left(\dfrac{r - 1}{2}, \dfrac{r - 1}{2}\right)^{\mathrm{th}}$ entry is

\[
(-1)^{\frac{r - 1}{2}}\frac{r + 3}{r + 1}.
\]

Since $p$ and $r$ are odd, we see that $r + 3 \neq p$. This shows $\det A \in \Zp^*$ and

\[
x_{\frac{r + 1}{2}} = (-1)^{\frac{r - 1}{2}}\frac{r + 1}{r + 3}.
\]

Now multiplying (<ref>) by $x_{n}$ and adding them up with $x_{\frac{r + 1}{2}}$ times (<ref>), we get

\[
0 \equiv \sum_{a = 0}^{p - 1}p^{0.5}(z - ap)^{\frac{r - 1}{2}}\mathbbm{1}_{ap + p^2\Zp} \mod \pi \latticeL{k}.
\]

Applying the matrix $\begin{pmatrix}p & 0 \\ 0 & 1\end{pmatrix}$, we get

\[
0 \equiv \sum_{a = 0}^{p - 1}(z - a)^{\frac{r - 1}{2}}\mathbbm{1}_{a + p\Zp} \mod \pi \latticeL{k}. \qedhere
\]

Now we turn to the claim made at the beginning of this section.

Let $1 \leq r \leq p-2$ be odd. For $\nu \geq 0.5$, set

\[
g(z) = p^{-1}\sum_{i = 0}^{p - 1}\lambda_i (z - i)^{\frac{r + 1}{2}}\logL(z - i),
\]

where the $\lambda_i \in \zp$ are as in Lemma <ref> (3).
Then, up to multiplication by $(-1)^{\frac{r-1}{2}} (\frac{r+1}{2})$,

\begin{eqnarray*}
g(z) & \equiv & \sum_{a = 0}^{p^2 - 1}p^{-1}(z - a)^{\frac{r + 1}{2}}\mathbbm{1}_{z + p^2\Zp} \\
&& + \sum_{a = 0}^{p - 1}(-1)^{\frac{r - 1}{2}}\frac{r + 1}{2}\left(\frac{\cL - H_{-} - H_{+}}{p^{0.5}}\right)p^{-0.5}(z - a)^{\frac{r + 1}{2}}\mathbbm{1}_{a + p\Zp} \\
&& + z^{\frac{r + 1}{2}}\mathbbm{1}_{\Zp} \mod \pi \latticeL{k}.
\end{eqnarray*}

Write $g(z) = g(z)\mathbbm{1}_{\Zp} + g(z)\mathbbm{1}_{\Qp \setminus \Zp}$. By Lemma <ref> (3)  $\lambda_0 = 1$ and $\lambda_i = 1 \mod p$ for $1 \leq i \leq p - 1$. Moreover,
the assumption $p \geq 5$ implies that $\sum_{i = 0}^{p - 1}\lambda_i i^j = 0$ for $0 \leq j \leq \dfrac{r + 1}{2} + 1 \leq p - 2$. As usual

\[
w \cdot g(z)\mathbbm{1}_{\Qp \setminus \Zp} = \sum_{i = 0}^{p - 1}p^{-1}\lambda_iz^{\frac{r - 1}{2}}(1 - zi)^{\frac{r + 1}{2}}\logL(1 - zi)\mathbbm{1}_{p\Zp}.
\]

Moreover, using Lemma <ref> (2),
this function vanishes modulo $\pi \latticeL{k}$. Therefore
$g(z)\mathbbm{1}_{\Qp \setminus \Zp} \equiv 0 \mod \pi \latticeL{k}$ and
$g(z) \equiv g(z)\mathbbm{1}_{\Zp} \mod \pi \latticeL{k}$.

Using Lemma <ref> (2), we see that $g(z)\mathbbm{1}_{\Zp} \equiv g_3(z) \mod \pi \latticeL{k}$. We analyze $g_3(z) - g_2(z)$, $g_2(z) - g_1(z)$ and $g_1(z)$ separately. First consider

\begin{eqnarray*}
g_3(z) - g_2(z) & = & \sum_{a = 0}^{p^2 - 1}\sum_{\alpha = 0}^{p - 1}\left[\sum_{j = 0}^{\frac{r - 1}{2}}\frac{g^{(j)}(a + \alpha p^2)}{j!}(z - a - \alpha p^2)^j\right]\mathbbm{1}_{a + \alpha p^2 + p^3\Zp} \\
&& - \sum_{a = 0}^{p^2 - 1}\left[\sum_{j = 0}^{\frac{r - 1}{2}}\frac{g^{(j)}(a)}{j!}(z - a)^j\right]\mathbbm{1}_{a + p^2\Zp}.
\end{eqnarray*}

The usual argument shows that

the coefficient of $(z - a - \alpha p^2)^j\mathbbm{1}_{a + \alpha p^2 + p^3\Zp}$ in $g_3(z) - g_2(z)$ is

\begin{eqnarray}\label{jth summand in g3 - g2 in main self-dual}
&& \!\!\!\! {\frac{r + 1}{2} \choose j}\sum_{i = 0}^{p - 1}p^{-1}\lambda_i\Big[(a + \alpha p^2 - i)^{\frac{r + 1}{2} - j}\logL(a + \alpha p^2 - i) - (a - i)^{\frac{r + 1}{2} - j}\logL(a - i) \\
&& \!\!\!\! - {\frac{r + 1}{2} - j \choose 1}(\alpha p^2)(a - i)^{\frac{r - 1}{2} - j}\logL(a - i) - \cdots - {\frac{r + 1}{2} - j \choose \frac{r - 1}{2} - j}(\alpha p^2)^{\frac{r - 1}{2} - j}(a - i)\logL(a - i)\Big]\nonumber.
\end{eqnarray}

First assume that $a \neq i$. Writing $\logL(a - i + \alpha p^2) = \logL(1 + (a - i)^{-1}\alpha p^2) + \logL(a - i)$, we see that the $i \neq a$ summand in expression (<ref>) is

\[
p^{-1}\lambda_i\left[(a - i + \alpha p^2)^{\frac{r + 1}{2} - j}\logL(1 + (a - i)^{-1}\alpha p^2) + (\alpha p^2)^{\frac{r + 1}{2} - j}\logL(a - i)\right].
\]

Expanding the first term in the display above, we see that the $i \neq a$ summand in (<ref>) is

\begin{eqnarray}\label{i neq a summands in g3 - g2 in self-dual}
&& p^{-1}\lambda_i\Big[(a - i)^{\frac{r - 1}{2} - j}\alpha p^2 + c_{\frac{r - 3}{2} - j}(a - i)^{\frac{r - 3}{2} - j}(\alpha p^2)^2 + \cdots + c_0(\alpha p^2)^{\frac{r + 1}{2} - j} \\
&& \qquad\qquad\qquad\qquad\qquad\qquad  + (\alpha p^2)^{\frac{r + 1}{2} - j}\logL(a - i)\Big] \mod p^{r - 2j}\pi, \nonumber
\end{eqnarray}

where $c_0, \> \ldots, \> c_{\frac{r - 3}{2} - j} \in \Zp$ and $c_0 = H_{\frac{r + 1}{2} - j}$.

Next assume that $i = a$. Then the $i^{\mathrm{th}}$ summand in expression (<ref>) is

\begin{eqnarray*}
p^{-1}\lambda_a(\alpha p^2)^{\frac{r + 1}{2} - j}\logL(\alpha p^2) = p^{-1}\lambda_a(\alpha p^2)^{\frac{r + 1}{2} - j}(2\cL) + p^{-1}\lambda_a(\alpha p^2)^{\frac{r + 1}{2} - j}\logL(\alpha).
\end{eqnarray*}

The second term on the right side of the equation above is $0$ modulo $p^{r - 2j}\pi$. Indeed, this is clear if $\alpha = 0$ and if $\alpha \neq 0$, then the valuation of the second term on the right side of the display above is $r + 1 - 2j > r - 2j$. Therefore the $i = a$ summand in expression (<ref>) is

\begin{eqnarray}\label{i = a summand in g3 - g2 in self-dual}
p^{-1}\lambda_a(\alpha p^2)^{\frac{r + 1}{2} - j}(2\cL) \mod p^{r - 2j}\pi.
\end{eqnarray}

If $a$ is in the range $[0, p - 1]$, then both the cases, $i \neq a$ and $i = a$ arise. If $a$ is in the range $[p, p^2 - 1]$, then only the first case, $i \neq a$ arises. Therefore if $0 \leq a \leq p - 1$, then using the identity $\sum_{i = 0}^{p - 1}\lambda_i i^{j} = 0$ for $0 \leq j \leq p - 2$ and Lemma <ref>, we see that the $a^{\mathrm{th}}$ summand in $g_3(z) - g_2(z)$ is

\begin{eqnarray}\label{0 leq a leq p - 1 summands in g3 - g2 in self-dual}
\!\!\sum_{\alpha = 0}^{p - 1}\sum_{j = 0}^{\frac{r - 1}{2}}\!{\frac{r + 1}{2} \choose j}p^{-1}\lambda_a\!\!\left(2\cL - H_{\frac{r + 1}{2} - j}\right)\!\!(\alpha p^2)^{\frac{r + 1}{2} - j}(z - a - \alpha p^2)^{j}\mathbbm{1}_{a + \alpha p^2 + p^3\Zp} \!\!\! \mod \pi \latticeL{k}.
\end{eqnarray}

Next if $p \leq a \leq p^2 - 1$, then expression (<ref>) is

\[
{\frac{r + 1}{2} \choose j}p^{-1}\lambda_{i_0}(\alpha p^2)^{\frac{r + 1}{2} - j}\logL(a - i_0) \mod p^{r - 2j}\pi,
\]

where $0 \leq i_0 \leq p - 1$ is the unique integer such that $i_0 \equiv a \!\!\mod p$. Since $0 \leq i_0 \leq p - 1$ and $p \leq a \leq p^2 - 1$, the valuation of the difference $a - i_0$ is $1$. So $\logL(a - i_0) = \cL \mod p\Zp$. This means that $\logL(a - i_0)$ in the expression above can be replaced by $\cL$ modulo $p^{r/2 - j}\pi$. Using Lemma <ref>, we therefore see that for $p \leq a \leq p^2 - 1$, the $a^{\mathrm{th}}$ summand in $g_3(z) - g_2(z)$ is

\begin{eqnarray}\label{p leq a leq p^2 - 1 summands in g3 - g2 in self-dual}
\sum_{\alpha = 0}^{p - 1}\sum_{j = 0}^{\frac{r - 1}{2}}{\frac{r + 1}{2} \choose j}p^{-1}\lambda_{i_0}\cL(\alpha p^2)^{\frac{r + 1}{2} - j}(z - a - \alpha p^2)^j\mathbbm{1}_{a + \alpha p^2 + p^3\Zp} \mod \pi \latticeL{k},
\end{eqnarray}

Combining expressions (<ref>) and (<ref>), using the fact that $\lambda_{i} \equiv 1 \mod p$ for $0 \leq i \leq p - 1$ and separating out one $\cL$ from expression (<ref>), we see that

\begin{eqnarray}\label{g3 - g2 in self-dual}
g_3(z) - g_2(z) \!\! & \equiv & \!\! \sum_{a = 0}^{p - 1}\sum_{\alpha = 0}^{p - 1}\sum_{j = 0}^{\frac{r - 1}{2}}{\frac{r + 1}{2} \choose j}p^{-1}\!\!\left(\cL - H_{\frac{r + 1}{2} - j}\right)(\alpha p^2)^{\frac{r + 1}{2} - j} % \\
%                 && \quad \quad \quad \quad \quad \quad \qquad
(z - a - \alpha p^2)^j\mathbbm{1}_{a + \alpha p^2 + p^3\Zp} \\ %\nonumber\\
&& \!\! + \!\! \sum_{a = 0}^{p^2 - 1}\!\sum_{\alpha = 0}^{p - 1}\!\sum_{j = 0}^{\frac{r - 1}{2}}\!{\frac{r + 1}{2} \choose j}p^{-1}\cL(\alpha p^2)^{\frac{r + 1}{2} - j}(z - a - \alpha p^2)^j\mathbbm{1}_{a + \alpha p^2 + p^3\Zp} \!\!\! \mod \pi \latticeL{k}. \nonumber
\end{eqnarray}

Now $\nu \geq 0.5$ implies that $\nu > 1 + r/2 - n$ for all $k/2 < n \leq r$, so by
Proposition <ref>, we get

\[
z^{r - n}\mathbbm{1}_{p \Zp} \equiv \sum_{b = 1}^{p - 1}\sum_{j = 1}^{\frac{r - 1}{2}}c(n, j)b^{r - j - n}(z - b)^j\mathbbm{1}_{b + p\Zp} \mod \pi \latticeL{k}.
\]

Applying the operator $\sum\limits_{\alpha = 0}^{p - 1}\alpha^{n - \frac{r - 1}{2}}\begin{pmatrix}1 & 0 \\ -a - \alpha p^2 & p^2\end{pmatrix}$ to both sides,

making the substitution $\alpha + b = \lambda$ on the right side and summing over $b$
using equation (<ref>) with $i = \dfrac{r + 1}{2}$,

replacing $\lambda$ by $\alpha$ and substituting the value of $c(n, j)$ obtained in Proposition <ref>,

we get

\begin{eqnarray}\label{Applied inductive steps for g3 - g2 in self-dual}
0 & \equiv & \sum_{\alpha = 0}^{p - 1}\sum_{j = r - n}^{\frac{r - 1}{2}}{n - \frac{r - 1}{2} \choose n - r + j}\frac{(r - n)! n}{(r - j) \cdots (n - j)}p^{-1}(\alpha p^2)^{\frac{r + 1}{2} - j} \\
&& \quad \quad \quad \quad \quad \quad \qquad (z - a - \alpha p^2)^j \mathbbm{1}_{a + \alpha p^2 + p^3\Zp} \mod \pi \latticeL{k} \nonumber
\end{eqnarray}

for each $k/2 < n \leq r$.

For a fixed $0 \leq a \leq p - 1$, we wish to write the $a^{\mathrm{th}}$ summand in the first term on the right side of equation (<ref>) as $x_n \in E$ times the right side of equation (<ref>) for $k/2 < n \leq r$ and add them to $x_{\frac{r + 1}{2}} \in E$ times

\[
\sum_{\alpha = 0}^{p - 1}p^{-1}(\alpha p^2)(z - a - \alpha p^2)^{\frac{r - 1}{2}}\mathbbm{1}_{a + \alpha p^2 + p^3\Zp}.
\]

Comparing coefficients of like terms of the form $p^{-1}(\alpha p^2)^{\frac{r + 1}{2} - j}(z - a - \alpha p^2)^j\mathbbm{1}_{a + \alpha p^2 + p^3\Zp}$, we get the following matrix equation

\[
\begin{pmatrix}
{\frac{r + 1}{2} \choose 0}\frac{r}{r} & 0 & 0 & \cdots & 0 \\
{\frac{r + 1}{2}  \choose 1}\frac{r}{r - 1} & {\frac{r - 1}{2} \choose 0}\frac{r - 1}{(r - 1)(r - 2)} & 0 & \cdots & 0 \\
{\frac{r + 1}{2} \choose 2}\frac{r}{r - 2} & {\frac{r - 1}{2} \choose 1}\frac{r - 1}{(r - 2)(r - 3)} & {\frac{r - 3}{2} \choose 0}\frac{2!(r - 2)}{(r - 2)(r - 3)(r - 4)} & \cdots & 0 \\
\vdots & \vdots & \vdots & & \vdots \\
{\frac{r + 1}{2} \choose \frac{r - 1}{2}}\frac{r}{\frac{r + 1}{2}} & {\frac{r - 1}{2} \choose \frac{r - 3}{2}}\frac{r - 1}{(\frac{r + 1}{2})(\frac{r - 1}{2})} & {\frac{r - 3}{2} \choose \frac{r - 5}{2}}\frac{2!(r - 2)}{(\frac{r + 1}{2})(\frac{r - 1}{2})(\frac{r - 3}{2})} & \cdots & 1
\end{pmatrix}
\begin{pmatrix}
x_r \\ x_{r - 1} \\ x_{r - 2} \\ \vdots \\ x_{\frac{r + 1}{2}}
\end{pmatrix}
\begin{pmatrix}
{\frac{r + 1}{2} \choose 0}(\cL - H_{\frac{r + 1}{2}}) \\
{\frac{r + 1}{2} \choose 1}(\cL - H_{\frac{r - 1}{2}}) \\
{\frac{r + 1}{2} \choose 2}(\cL - H_{\frac{r - 3}{2}}) \\
\vdots \\
{\frac{r + 1}{2} \choose \frac{r - 1}{2}}(\cL - H_{1})
\end{pmatrix}.
\]

It can be checked[See footnote <ref>.]
that $x_{\frac{r + 1}{2}} = (-1)^{\frac{r - 1}{2}}\left(\cL - H_{-} - H_{+}\right)$.

Therefore, we can write equation (<ref>) as

\begin{eqnarray*}
g_3(z) - g_2(z) & \equiv & \sum_{a = 0}^{p - 1}\sum_{\alpha = 0}^{p - 1}(-1)^{\frac{r - 1}{2}}\left(\cL - H_{-} - H_{+}\right)p^{-1}(\alpha p^2)(z - a - \alpha p^2)^{\frac{r - 1}{2}}\mathbbm{1}_{a + \alpha p^2 + p^3\Zp} \\
&& + \sum_{a = 0}^{p^2 - 1}\sum_{\alpha = 0}^{p - 1}\sum_{j = 0}^{\frac{r - 1}{2}}{\frac{r + 1}{2} \choose j}p^{-1}\cL(\alpha p^2)^{\frac{r + 1}{2} - j}(z - a - \alpha p^2)^j\mathbbm{1}_{a + \alpha p^2 + p^3\Zp} \\
&& \qquad \qquad \mod \pi \latticeL{k}.
\end{eqnarray*}

Since $\nu \geq 0.5$, the first term on the right side of the display above is $0$ modulo $\pi \latticeL{k}$ by Lemma <ref>. Next, we simplify the second line on the right side of the display above using the binomial theorem to get

\[
g_3(z) - g_2(z) \equiv \sum_{a = 0}^{p^2 - 1}\sum_{\alpha = 0}^{p - 1}p^{-1}\cL\left[(z - a)^\frac{r + 1}{2} - (z - a - \alpha p^2)^{\frac{r + 1}{2}}\right]\mathbbm{1}_{a + \alpha p^2 + p^3\Zp} \mod \pi \latticeL{k}.
\]

Note that $\nu \geq 0.5 \geq 0$ implies $\cL \in \Zp$. Therefore using Lemma <ref> the second term on the right side of the display above is $0$ modulo $\pi \latticeL{k}$. Since the first term on the right side of the display above does not involve $\alpha$, the characteristic functions sum up to $\mathbbm{1}_{a + p^2\Zp}$. Therefore

\begin{eqnarray}\label{Final g3 - g2 in self-dual}
g_3(z) - g_2(z) \equiv \sum_{a = 0}^{p^2 - 1}p^{-1}\cL(z - a)^{\frac{r + 1}{2}}\mathbbm{1}_{a + p^2\Zp} \mod \pi \latticeL{k}.
\end{eqnarray}

We now simplify $g_2(z) - g_1(z)$.
The coefficient of $(z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2\Zp}$ in $g_2(z) - g_1(z)$ is

\begin{eqnarray}\label{jth summand in g2 - g1 in main self-dual}
&& \!\!\!\!\!\!\!\! {\frac{r + 1}{2} \choose j}\sum_{i = 0}^{p - 1}p^{-1}\lambda_i\Big[(a + \alpha p - i)^{\frac{r + 1}{2} - j}\logL(a + \alpha p - i)^{\frac{r + 1}{2} - j} - (a - i)^{\frac{r + 1}{2} - j}\logL(a - i) \\
&& \!\!\!\!\!\!\!\! - {\frac{r + 1}{2} - j \choose 1}(\alpha p)(a - i)^{\frac{r - 1}{2} - j}\logL(a - i) - \cdots - {\frac{r + 1}{2} - j \choose \frac{r - 1}{2} - j}(\alpha p)^{\frac{r - 1}{2} - j}(a - i)\logL(a - i)\Big]. \nonumber
\end{eqnarray}

Analyzing the $i \neq a$ and $i = a$ terms separately and

First assume that $i \neq a$. Writing $\logL(a - i + \alpha p) = \logL(1 + (a - i)^{-1}\alpha p) + \logL(a - i)$, we see that the $i \neq a$ summand in expression (<ref>) is

\[
p^{-1}\lambda_i\left[(a - i + \alpha p)^{\frac{r + 1}{2} - j}\logL(1 + (a - i)^{-1}\alpha p) + (\alpha p)^{\frac{r + 1}{2} - j}\logL(a - i)\right].
\]

As $0 \leq a, \> i \leq p - 1$, the condition $a \neq i$ implies that $p \nmid a - i$. Therefore $p \mid \logL(a - i)$. Moreover, we expand the first term in the display above to see that the $i \neq a$ summand in expression (<ref>) is

\begin{eqnarray}\label{i neq a summand in g2 - g1 in main self-dual}
\!\!\!\! p^{-1}\lambda_i\left[(a - i)^{\frac{r - 1}{2} - j}(\alpha p) + c_{\frac{r - 3}{2} - j}(a - i)^{\frac{r - 3}{2} - j}(\alpha p)^2 + \cdots + c_0(\alpha p)^{\frac{r + 1}{2} - j}\right] \!\!\! \mod p^{r/2 - j}\pi,
\end{eqnarray}

where $c_{0}, \> \ldots, \> c_{\frac{r - 3}{2} - j} \in \Zp$ and $c_0 = H_{\frac{r + 1}{2} - j}$.

Next, assume that $i = a$. The $i^{\mathrm{th}}$ summand in expression (<ref>) is

\[
p^{-1}\lambda_a(\alpha p)^{\frac{r + 1}{2} - j}\logL(\alpha p) = p^{-1}\lambda_a(\alpha p)^{\frac{r + 1}{2} - j}\cL + p^{-1}\lambda_a(\alpha p)^{\frac{r + 1}{2} - j}\logL(\alpha).
\]

The second term in the display above is $0$ modulo $p^{r/2 - j}\pi$. Indeed, this is clear if $\alpha = 0$ and if $\alpha \neq 0$, then the valuation of the second term in the display above is $(r + 1)/2 - j > r/2 - j$. So the $i = a$ summand in expression (<ref>) is

\begin{eqnarray}\label{i = a summand in g2 - g1 in main self-dual}
p^{-1}\lambda_a(\alpha p)^{\frac{r + 1}{2} - j}\cL \mod p^{r/2 - j}\pi.
\end{eqnarray}

Putting expressions (<ref>) and (<ref>) in expression (<ref>) and using $\sum_{i = 0}^{p - 1}\lambda_i i^{j} = 0$ for $0 \leq j \leq p - 2$, we see that the coefficient of $(z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2\Zp}$ in $g_2(z) - g_1(z)$ is

\begin{eqnarray*}
p^{-1}\lambda_a\left(\cL - H_{\frac{r + 1}{2} - j}\right)(\alpha p)^{\frac{r + 1}{2} - j} \mod p^{r/2 - j}\pi.
\end{eqnarray*}

using $\sum_{i = 0}^{p - 1}\lambda_i i^{j} = 0$ for $0 \leq j \leq p - 2$,

we get as usual

\begin{eqnarray}\label{g2 - g1 in self-dual}
g_2(z) - g_1(z) & \equiv & \sum_{a = 0}^{p - 1}\sum_{\alpha = 0}^{p - 1}\sum_{j = 0}^{\frac{r - 1}{2}}{\frac{r + 1}{2} \choose j}p^{-1}\left(\cL - H_{\frac{r + 1}{2} - j}\right)(\alpha p)^{\frac{r + 1}{2} - j} \\
&& \quad \quad \quad \quad \quad \quad \qquad (z - a - \alpha p)^{j}\mathbbm{1}_{a + \alpha p + p^2 \Zp} \mod \pi \latticeL{k}. \nonumber
\end{eqnarray}

Assume momentarily that $r > 1$. Proposition <ref> gives us

\begin{eqnarray}\label{Applied hard inductive steps in g2 - g1 in self-dual}
0 & \equiv & \sum_{\alpha = 0}^{p - 1}\sum_{j = r-n}^{{\frac{r+1}{2}}}p^{\frac{r-1}{2} - j}{n - \frac{r-1}{2} \choose n - r + j}\frac{(r-n)! n}{(r - j)\cdots (n - j)}\alpha^{\frac{r+1}{2} - j} \\
&& \quad \quad \quad \quad \quad \quad \qquad (z - a - \alpha p)^j \mathbbm{1}_{a + \alpha p + p^2\Zp} \mod \pi \latticeL{k} \nonumber
\end{eqnarray}

for every $\dfrac{r + 3}{2} \leq n \leq r$. For each $a$ and $\alpha$ in $[0, p - 1]$, we wish to write the coefficient of $p^{-1}(\alpha p)^{\frac{r + 1}{2} - j}(z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2\Zp}$ in equation (<ref>) as a sum of $x_n \in E$ times the corresponding coefficient in equation (<ref>), $x_{\frac{r + 1}{2}} \in E$ times

\[
\sum_{\alpha = 0}^{p - 1}\alpha (z - a - \alpha p)^{\frac{r - 1}{2}}\mathbbm{1}_{a + \alpha p + p^2 \Zp}
\]

and $x_{\frac{r - 1}{2}} \in E$ times

\[
\sum_{\alpha = 0}^{p - 1}p^{-1}(z - a - \alpha p)^{\frac{r + 1}{2}}\mathbbm{1}_{a + \alpha p + p^2\Zp}.
\]

Therefore we get the following matrix equation

\begin{eqnarray}\label{First matrix equation in g2 - g1 in self-dual}
\!\!\!\! \tiny\begin{pmatrix}
{\frac{r + 1}{2} \choose 0}\frac{r}{r} & 0 & 0 & \cdots & 0 & 0\\
{\frac{r + 1}{2}  \choose 1}\frac{r}{r - 1} & {\frac{r - 1}{2} \choose 0}\frac{r - 1}{(r - 1)(r - 2)} & 0 & \cdots & 0 & 0\\
{\frac{r + 1}{2} \choose 2}\frac{r}{r - 2} & {\frac{r - 1}{2} \choose 1}\frac{r - 1}{(r - 2)(r - 3)} & {\frac{r - 3}{2} \choose 0}\frac{2!(r - 2)}{(r - 2)(r - 3)(r - 4)} & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & & \vdots & \vdots\\
{\frac{r + 1}{2} \choose \frac{r - 1}{2}}\frac{r}{\frac{r + 1}{2}} & {\frac{r - 1}{2} \choose \frac{r - 3}{2}}\frac{r - 1}{(\frac{r + 1}{2})(\frac{r - 1}{2})} & {\frac{r - 3}{2} \choose \frac{r - 5}{2}}\frac{2!(r - 2)}{(\frac{r + 1}{2})(\frac{r - 1}{2})(\frac{r - 3}{2})} & \cdots & 1 & 0\\
{\frac{r + 1}{2} \choose \frac{r + 1}{2}}\frac{r}{\frac{r - 1}{2}} & {\frac{r - 1}{2} \choose \frac{r - 1}{2}}\frac{r - 1}{(\frac{r - 1}{2})(\frac{r - 3}{2})} & {\frac{r - 3}{2} \choose \frac{r - 3}{2}}\frac{2!(r - 2)}{(\frac{r - 1}{2})(\frac{r - 3}{2})(\frac{r - 5}{2})} & \cdots & 0 & 1
\end{pmatrix} \!\!
\begin{pmatrix}
x_r \\ x_{r - 1} \\ x_{r - 2} \\ \vdots \\ x_{\frac{r + 1}{2}} \\ x_{\frac{r - 1}{2}}
\end{pmatrix} \!\!
= \!\!
\begin{pmatrix}
{\frac{r + 1}{2} \choose 0}(\cL - H_{\frac{r + 1}{2}}) \\
{\frac{r + 1}{2} \choose 1}(\cL - H_{\frac{r - 1}{2}}) \\
{\frac{r + 1}{2} \choose 2}(\cL - H_{\frac{r - 3}{2}}) \\
\vdots \\
{\frac{r + 1}{2} \choose \frac{r - 1}{2}}(\cL - H_{1}) \\
\end{pmatrix}.
\end{eqnarray}

It can be checked[See footnote <ref>.] that

\[
x_{\frac{r + 1}{2}} = (-1)^{\frac{r - 1}{2}}\left(\cL - H_{-} - H_{+}\right) \text{ and } x_{\frac{r - 1}{2}} \equiv -\cL + (-1)^{\frac{r - 1}{2}}\frac{1}{\frac{r + 1}{2}} \mod \pi.
\]

Using Lemma <ref>,
we can therefore write equation (<ref>) as

\begin{eqnarray}\label{Intermediate g2 - g1 in self-dual}
\!\!\!\!\!\! g_2(z) - g_1(z) \!\! & \equiv & \!\! \sum_{a = 0}^{p - 1}\sum_{\alpha = 0}^{p - 1}(-1)^{\frac{r - 1}{2}}\!\left(\frac{\cL - H_{-} - H_{+}}{p^{0.5}}\right)\!p^{0.5}\alpha (z - a - \alpha p)^{\frac{r - 1}{2}}\mathbbm{1}_{a + \alpha p + p^2\Zp} \\
&& \!\! + \sum_{a = 0}^{p - 1}\sum_{\alpha = 0}^{p - 1}\left(-\cL + (-1)^{\frac{r - 1}{2}}\frac{1}{\frac{r + 1}{2}}\right)p^{-1}(z - a - \alpha p)^{\frac{r + 1}{2}}\mathbbm{1}_{a + \alpha p + p^2\Zp} \nonumber \\
&& \!\! \qquad \qquad \mod \pi \latticeL{k}. \nonumber
\end{eqnarray}

The expressions (<ref>) and (<ref>) coincide
if $r = 1$ as well using Lemma <ref>.
As $0 \leq a, \> \alpha \leq p - 1$, we see that $a + \alpha p$ varies from $0$ to $p^2 - 1$. Therefore we replace $a + \alpha p$ by $a$ varying between $0$ and $p^2 - 1$ in the second term on the right side of the equation (<ref>).

Fixing $0 \leq a \leq p - 1$, we write the $a^{\mathrm{th}}$ summand in the first term on the right side of the equation (<ref>) as a multiple of

\[
p^{-0.5}(z - a)^{\frac{r + 1}{2}}\mathbbm{1}_{a + p\Zp}.
\]

We first expand

\begin{eqnarray}\label{Binomial equation in inductive steps for g2 - g1 in self-dual}
\!\!\!\!\!\! p^{-0.5}(z - a)^{\frac{r + 1}{2}}\mathbbm{1}_{a + p\Zp} & \equiv & \sum_{\alpha = 0}^{p - 1}\sum_{j = 0}^{\frac{r - 1}{2}}{\frac{r + 1}{2} \choose j}p^{-0.5}(\alpha p)^{\frac{r + 1}{2} - j}(z - a - \alpha p)^j \mathbbm{1}_{a + \alpha p + p^2\Zp} \\
&& \qquad \qquad \mod \pi \latticeL{k}. \nonumber
\end{eqnarray}

Note that we have dropped the $j = \dfrac{r + 1}{2}$ term using Lemma <ref>. Since $\nu \geq 0.5$, using Proposition <ref> we have the following shallow inductive steps

\[
z^{r - n}\mathbbm{1}_{p\Zp} \equiv \sum_{b = 1}^{p - 1}\sum_{j = 1}^{\frac{r - 1}{2}}c(n, j)b^{r - j - n}(z - b)^j\mathbbm{1}_{b + p\Zp} \mod \pi \latticeL{k}
\]

for each $\dfrac{r + 3}{2} \leq n \leq r$. Applying $\sum\limits_{\alpha = 0}^{p - 1}\alpha^{n - \frac{r - 1}{2}}\begin{pmatrix}1 & 0 \\ - a - \alpha p & p\end{pmatrix}$ to both sides,

substituting $\lambda = \alpha + b$ on the right side and summing over $b$ using equation (<ref>)
with $i = \dfrac{r + 1}{2}$,

replacing $\lambda$ by $\alpha$, and substituting the value of $c(n, j)$ from Proposition <ref>,

we get

\begin{eqnarray}\label{Applied shallow inductive steps in g2 - g1 in self-dual}
0 & \equiv & \sum_{\alpha = 0}^{p - 1}\sum_{j = r - n}^{\frac{r - 1}{2}}{n - \frac{r - 1}{2} \choose n - r + j}\frac{(r - n)!n}{(r - j) \cdots (n - j)}p^{-0.5}(\alpha p)^{\frac{r + 1}{2} - j} \\
&& \qquad \qquad \qquad \qquad (z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2\Zp} \mod \pi \latticeL{k}. \nonumber
\end{eqnarray}

Now write the coefficient of $p^{-0.5}(\alpha p)^{\frac{r + 1}{2} - j}(z - a - \alpha p)^j\mathbbm{1}_{a + \alpha p + p^2\Zp}$ in the first summand on the right side of equation (<ref>) as a sum of $x_n \in E$ times the corresponding coefficient in equation (<ref>) and $x_{\frac{r + 1}{2}} \in E$ times the corresponding coefficient in (<ref>). Therefore we get the following matrix equation

\[
{\frac{r + 1}{2} \choose 0}\frac{r}{r} & 0 & 0 & \cdots & {\frac{r + 1}{2} \choose 0} \\
{\frac{r + 1}{2}  \choose 1}\frac{r}{r - 1} & {\frac{r - 1}{2} \choose 0}\frac{r - 1}{(r - 1)(r - 2)} & 0 & \cdots & {\frac{r + 1}{2} \choose 1} \\
{\frac{r + 1}{2} \choose 2}\frac{r}{r - 2} & {\frac{r - 1}{2} \choose 1}\frac{r - 1}{(r - 2)(r - 3)} & {\frac{r - 3}{2} \choose 0}\frac{2!(r - 2)}{(r - 2)(r - 3)(r - 4)} & \cdots & {\frac{r + 1}{2} \choose 2} \\
\vdots & \vdots & \vdots & & \vdots \\
{\frac{r + 1}{2} \choose \frac{r - 1}{2}}\frac{r}{\frac{r + 1}{2}} & {\frac{r - 1}{2} \choose \frac{r - 3}{2}}\frac{r - 1}{(\frac{r + 1}{2})(\frac{r - 1}{2})} & {\frac{r - 3}{2} \choose \frac{r - 5}{2}}\frac{2!(r - 2)}{(\frac{r + 1}{2})(\frac{r - 1}{2})(\frac{r - 3}{2})} & \cdots & {\frac{r + 1}{2} \choose \frac{r - 1}{2}}
\end{pmatrix}
\begin{pmatrix}
x_r \\ x_{r - 1} \\ x_{r - 2} \\ \vdots \\ x_{\frac{r + 1}{2}}
\end{pmatrix}
\begin{pmatrix}
0 \\ 0 \\ 0 \\ \vdots \\ (-1)^{\frac{r - 1}{2}}\left(\frac{\cL - H_{-} - H_{+}}{p^{0.5}}\right)
\end{pmatrix}.}
\]

The determinant of this matrix is the same as that of the matrix $A$ in the matrix equation
(<ref>) with $i$ there equal to $\dfrac{r+1}{2}$.
By Cramer's rule, we see $x_{\frac{r + 1}{2}} = \dfrac{\cL - H_{-} - H_{+}}{p^{0.5}}$. We finally get

\begin{eqnarray}\label{Final g2 - g1 in self-dual}
g_2(z) - g_1(z) & \equiv & \sum_{a = 0}^{p - 1}\left(\frac{\cL - H_{-} - H_{+}}{p^{0.5}}\right)p^{-0.5}(z - a)^{\frac{r + 1}{2}}\mathbbm{1}_{a + p\Zp} \\
&& + \sum_{a = 0}^{p^2 - 1}\left(-\cL + (-1)^{\frac{r - 1}{2}}\frac{1}{\frac{r + 1}{2}}\right)p^{-1}(z - a)^{\frac{r + 1}{2}}\mathbbm{1}_{a + p^2\Zp} \mod \pi \latticeL{k}. \nonumber
\end{eqnarray}

Next we simplify $g_1(z)$. Recall that

\begin{equation}\label{g1 in self-dual}
g_1(z) = \sum_{a = 0}^{p - 1}\sum_{j = 0}^{\frac{r - 1}{2}}\frac{g^{(j)}(a)}{j!}(z - a)^j\mathbbm{1}_{a + p\Zp},
\end{equation}

\begin{eqnarray}\label{jth summand in g1 in self-dual}
\frac{g^{(j)}(a)}{j!} = {\frac{r + 1}{2} \choose j}\sum_{i = 0}^{p - 1}\frac{\lambda_i}{p}(a - i)^{\frac{r + 1}{2} - j}\logL(a - i).
\end{eqnarray}

The $i = a$ summand is $0$. For $i \neq a$, we write $a - i = [a - i] + pa_i$ for some $a_i \in \Zp$. Since $[a - i]$ is a root of unity, $\logL(a - i) = \logL(1 + [a - i]^{-1}pa_i)$. Expanding this logarithm, we get

\begin{eqnarray*}
\frac{g^{(j)}(a)}{j!} & \equiv & {\frac{r + 1}{2} \choose j}\sum_{\substack{i = 0 \\ i \neq a}}^{p - 1}\frac{\lambda_i}{p}[a - i]^{\frac{r - 1}{2} - j}pa_i \mod \pi \\
& \equiv & {\frac{r + 1}{2} \choose j}\sum_{\substack{i = 0 \\ i \neq a}}^{p - 1}\frac{\lambda_i}{p}\left(\frac{(a - i)^{\frac{r + 1}{2} - j} - [a - i]^{\frac{r + 1}{2} - j}}{\frac{r + 1}{2} - j}\right) \mod \pi.
\end{eqnarray*}

The expression inside the parentheses is divisible by $p$. Since $\lambda_i \equiv 1 \mod p$, we may replace $\lambda_i$ in the last line above by $1$. Moreover, since $0 \leq j \leq \dfrac{r - 1}{2}$, the power $\dfrac{r + 1}{2} - j$ is in the range $[1, p - 2]$. Using equation (<ref>), we see that the equation above simplifies to

\[
\frac{g^{(j)}(a)}{j!} \equiv {\frac{r + 1}{2} \choose j}\sum_{i = 0}^{p - 1}\frac{1}{p}\frac{(a - i)^{\frac{r + 1}{2} - j}}{\frac{r + 1}{2} - j} \equiv {\frac{r + 1}{2} \choose j}\sum_{m = 0}^{\frac{r + 1}{2} - j}{\frac{r + 1}{2} - j \choose m}(-1)^m \frac{S_m}{p}\frac{a^{\frac{r + 1}{2} - j - m}}{\frac{r + 1}{2} - j} \mod \pi.
\]

Putting this in equation (<ref>), and using Lemma <ref>, we get

\begin{eqnarray*}
g_1(z) \equiv \sum_{a = 0}^{p - 1}\sum_{j = 0}^{\frac{r - 1}{2}}\sum_{m = 0}^{\frac{r + 1}{2} - j}{\frac{r + 1}{2} \choose j}{\frac{r + 1}{2} - j \choose m}(-1)^m\frac{S_m}{p}\frac{a^{\frac{r + 1}{2} - j - m}}{\frac{r + 1}{2} - j}(z - a)^j\mathbbm{1}_{a + p\Zp} \mod \pi \latticeL{k}.
\end{eqnarray*}

The $m = 0$ part of this sum will be the important part of $g_1(z)$. We separate out the
$m \neq 0$ terms
and interchange the order of the $m$ and $j$ sums to get

\begin{eqnarray}\label{Intermediate g1 in self-dual}
g_1(z) \!\! & \equiv & \!\! \sum_{a = 0}^{p - 1}\sum_{j = 0}^{\frac{r - 1}{2}}{\frac{r + 1}{2} \choose j}\frac{a^{\frac{r + 1}{2} - j}}{\frac{r + 1}{2} - j}(z - a)^j\mathbbm{1}_{a + p\Zp} \\
&& \!\! + \sum_{a = 0}^{p - 1}\sum_{m = 1}^{\frac{r + 1}{2}}\sum_{j = 0}^{\frac{r + 1}{2} - m}\!\!{\frac{r + 1}{2} \choose j}{\frac{r + 1}{2} - j \choose m}(-1)^m \frac{S_m}{p}\frac{a^{\frac{r + 1}{2} - j - m}}{\frac{r + 1}{2} - j}(z - a)^j \mathbbm{1}_{a + p\Zp} \!\!\! \mod \pi \latticeL{k}. \nonumber
\end{eqnarray}

We show the second sum on the right side of (<ref>) is $0$ mod $\pi \latticeL{k}$. Fix an $m$ in $\left[1, \dfrac{r - 1}{2}\right]$. Using the shallow inductive steps
Proposition <ref> for $n = r, \> r - 1, \ldots, \dfrac{r + 1}{2} + m$, we get

\[
z^{r - n}\mathbbm{1}_{p\Zp} \equiv \sum_{b = 1}^{p - 1}\sum_{j = 1}^{\frac{r - 1}{2}}c(n, j)b^{r - j - n}(z - b)^j\mathbbm{1}_{b + p\Zp} \mod \pi \latticeL{k}.
\]

Applying the operator $\sum\limits_{a = 0}^{p - 1}a^{n - \frac{r - 1}{2} - m}\begin{pmatrix}1 & 0 \\ -a & 1\end{pmatrix}$, we get

\begin{eqnarray*}
\sum_{a = 0}^{p - 1}a^{n - \frac{r - 1}{2} - m}(z - a)^{r - n}\mathbbm{1}_{a + p\Zp} \!\! & \equiv & \!\! \sum_{a = 0}^{p - 1}\sum_{b = 1}^{p - 1}\sum_{j = 1}^{\frac{r - 1}{2}}a^{n - \frac{r - 1}{2} - m}b^{r - j - n}c(n, j)(z - (a + b))^j\mathbbm{1}_{(a + b) + p\Zp} \\
&& \!\! \qquad \qquad \mod \pi \latticeL{k}.
\end{eqnarray*}

Substituting $\lambda = a + b$ and summing over $b$ using equation (<ref>) with
$i = \dfrac{r+1}{2} + m$, we get

\begin{eqnarray*}
\sum_{a = 0}^{p - 1}a^{n - \frac{r - 1}{2} - m}(z - a)^{r - n}\mathbbm{1}_{a + p\Zp} & \equiv & \sum_{\lambda = 0}^{p - 1}\sum_{j = r - n}^{\frac{r - 1}{2}}(-1)^{n - r + j + 1}{n - \frac{r - 1}{2} - m \choose n - r + j}c(n, j)\lambda^{\frac{r + 1}{2} - m - j} \\
&& \quad \quad \quad \quad \quad \quad (z - \lambda)^j\mathbbm{1}_{\lambda + p\Zp} \mod \pi \latticeL{k}.
\end{eqnarray*}

For $j > \dfrac{r + 1}{2} - m$, the binomial coefficient is $0$. Therefore we reduce the upper limit on $j$ to $\dfrac{r + 1}{2} - m$. Note that $m \geq 1$ implies that $\dfrac{r + 1}{2} - m \leq \dfrac{r - 1}{2}$. We replace $\lambda$ in the equation above by $a$ and substitute the value of $c(n, j)$ obtained in Proposition <ref>. The $-1$ part of $c(n, l)$ cancels with the term on the left side. Therefore we get

\begin{eqnarray}\label{Refined inductive steps to bring the power to (r+1)/2 - m}
\!\!\! 0 \equiv \! \sum_{a = 0}^{p - 1}\sum_{j = r - n}^{\frac{r + 1}{2} - m}\!\!\!{n - \frac{r - 1}{2} - m \choose n - r + j}\!\frac{(r - n)! n}{(r - j) \cdots (n - j)}a^{\frac{r + 1}{2} - m - j}(z - a)^j\mathbbm{1}_{a + p\Zp} \!\!\!\! \mod \pi \latticeL{k}.
\end{eqnarray}

We wish to write the term $a^{\frac{r + 1}{2} - j - m}(z - a)^j\mathbbm{1}_{a + p\Zp}$ in the second sum on the right side of (<ref>) as a sum of $x_n \in E$ times the corresponding term in equation (<ref>) for $\dfrac{r+1}{2} + m \leq n \leq r$
and $x_{\frac{r - 1}{2} + m} \in E$ times

\[
(z - a)^{\frac{r + 1}{2} - m}\mathbbm{1}_{a + p\Zp}.
\]

To this end, consider the matrix equation

\[
\begin{pmatrix}
{\frac{r + 1}{2} - m \choose 0}\frac{r}{r} & 0 & \cdots & 0 \\
{\frac{r + 1}{2} - m \choose 1}\frac{r}{r - 1} & {\frac{r - 1}{2} - m \choose 0}\frac{r - 1}{(r - 1)(r - 2)} & \cdots & 0 \\
{\frac{r + 1}{2} - m \choose 2}\frac{r}{r - 2} & {\frac{r - 1}{2} - m \choose 1}\frac{r - 1}{(r - 2)(r - 3)} & \cdots & 0 \\
\vdots & \vdots & & \vdots \\
{\frac{r + 1}{2} - m \choose \frac{r + 1}{2} - m}\frac{r}{\frac{r - 1}{2} + m} & {\frac{r - 1}{2} + m \choose \frac{r - 1}{2} + m}\frac{r - 1}{(\frac{r - 1}{2} + m)(\frac{r - 3}{2} + m)} & \cdots & 1
\end{pmatrix} \!\!
\begin{pmatrix}
x_r \\ x_{r - 1} \\ \vdots \\ x_{\frac{r - 1}{2} + m}
\end{pmatrix} \!\!
= \!\! \begin{pmatrix}
{\frac{r + 1}{2} \choose 0}{\frac{r + 1}{2} \choose m}(-1)^m\frac{S_m}{p}\frac{1}{\frac{r + 1}{2}} \\
{\frac{r + 1}{2} \choose 1}{\frac{r - 1}{2} \choose m}(-1)^m\frac{S_m}{p}\frac{1}{\frac{r - 1}{2}} \\
{\frac{r + 1}{2} \choose 2}{\frac{r - 3}{2} \choose m}(-1)^m\frac{S_m}{p}\frac{1}{\frac{r - 3}{2}} \\ \vdots \\ {\frac{r + 1}{2} \choose \frac{r + 1}{2} - m}{m \choose m}(-1)^m\frac{S_m}{p}\frac{1}{m}
\end{pmatrix}\!\!.
\]

Since the determinant of the coefficient matrix is a $p$-adic unit, there are solutions $x_{n} \in \Zp$ for $\dfrac{r - 1}{2} + m \leq n \leq r$. This shows that for $m \in \left[1, \dfrac{r - 1}{2}\right]$, the $m^{\mathrm{th}}$ summand in equation (<ref>) is

\begin{eqnarray}\label{First step in killing nonzero m terms}
x_{\frac{r - 1}{2} + m}\sum_{a = 0}^{p - 1}(z - a)^{\frac{r + 1}{2} - m}\mathbbm{1}_{a + p\Zp} \mod \pi \latticeL{k} \nonumber
\end{eqnarray}

for an integer $x_{\frac{r - 1}{2} + m}$. This statement is clearly also true for $m = \dfrac{r + 1}{2}$ since then $j$ can only take the value $0 = \dfrac{r + 1}{2} - m$.

We claim that

\[
\sum_{a = 0}^{p - 1} (z - a)^{\frac{r + 1}{2} - m}\mathbbm{1}_{a + p\Zp} \equiv 0 \mod \pi \latticeL{k}.
\]

This is clear for $m = 1$ by Lemma <ref>. So assume that $m$ belongs to
$\left[2, \dfrac{r + 1}{2}\right]$. Again using the shallow inductive step in
Proposition <ref>
for $n = \frac{r - 1}{2} + m$, we get

\[
z^{r - n}\mathbbm{1}_{p\Zp} \equiv \sum_{b = 1}^{p - 1}\sum_{j = 1}^{\frac{r - 1}{2}}c(n, j)b^{r - j - n}(z - b)^j\mathbbm{1}_{b + p\Zp} \mod \pi \latticeL{k}.
\]

Applying the operator $\sum\limits_{a = 0}^{p - 1}a^{n - \frac{r - 1}{2} - m}\begin{pmatrix}1 & 0 \\ - a & 1\end{pmatrix}$ on both sides,

substituting $\lambda = a + b$ and summing over $b$ on the right side
using equation (<ref>)
with $i = \dfrac{r+1}{2}$, we get as just seen above

\begin{eqnarray*}
\sum_{a = 0}^{p - 1}a^{n - \frac{r - 1}{2} - m}(z - a)^{r - n}\mathbbm{1}_{a + p\Zp} & \equiv & \sum_{\lambda = 0}^{p - 1}\sum_{j = r - n}^{\frac{r - 1}{2}}(-1)^{n - r + j + 1}{n - \frac{r - 1}{2} - m \choose n - r + j}c(n, j)\lambda^{\frac{r + 1}{2} - m - j} \\
&& \quad \quad \quad \quad \quad \quad(z - \lambda)^j\mathbbm{1}_{\lambda + p\Zp} \mod \pi \latticeL{k}.
\end{eqnarray*}

If $j > \dfrac{r + 1}{2} - m$, then the binomial coefficient is $0$. So we reduce the upper limit on $j$ to $\dfrac{r + 1}{2} - m$. Moreover, the lower limit on $j$ is $r - n = \dfrac{r + 1}{2} - m$. Therefore we get

\begin{eqnarray*}
\sum_{a = 0}^{p - 1}a^{n - \frac{r - 1}{2} - m}(z - a)^{r - n}\mathbbm{1}_{a + p\Zp} & \equiv & \sum_{\lambda = 0}^{p - 1}-c\left(\frac{r - 1}{2} + m, \frac{r + 1}{2} - m\right)(z - \lambda)^{\frac{r + 1}{2} - m}\mathbbm{1}_{\lambda + p\Zp} \\
&& \qquad \qquad \mod \pi \latticeL{k}.
\end{eqnarray*}

We replace $\lambda$ by $a$. Putting the value of $c\left(\frac{r - 1}{2} + m, \frac{r + 1}{2} - m\right)$ and cancelling the $-1$ part with the left side, we get

\[
0 \equiv \sum_{a = 0}^{p - 1}\frac{\left(\frac{r + 1}{2} - m\right)!\left(\frac{r - 1}{2} + m\right)}{\left(\frac{r - 1}{2} + m\right)\cdots\left(2m - 1\right)}(z - a)^{\frac{r + 1}{2} - m}\mathbbm{1}_{a + p\Zp} \mod \pi \latticeL{k}.
\]

Now $2 \leq m \leq \dfrac{r + 1}{2}$ implies that the coefficient in the equation above is a unit.
This proves the claim for such $m$ as well.

Thus, the second sum on the right side of (<ref>)
is $0$ mod $\pi \latticeL{k}$.

We get

\begin{eqnarray}
\label{Penultimate g1 in self-dual}
g_1(z) \equiv \sum_{a = 0}^{p - 1}\sum_{j = 0}^{\frac{r - 1}{2}}{\frac{r + 1}{2} \choose j}\frac{a^{\frac{r + 1}{2} - j}}{\frac{r + 1}{2} - j}(z - a)^j\mathbbm{1}_{a + p\Zp} \mod \pi \latticeL{k}.
\end{eqnarray}

We wish to write all the powers of $(z - a)$ appearing on the right side of the equation above in terms of $(z - a)^{\frac{r + 1}{2}}$. So for $\dfrac{r + 3}{2} \leq n \leq r$, consider again
the shallow inductive steps

\[
z^{r - n}\mathbbm{1}_{p\Zp} \equiv \sum_{b = 1}^{p - 1}\sum_{j = 1}^{\frac{r - 1}{2}}c(n, j)b^{r - j - n}(z - b)^j\mathbbm{1}_{b + p\Zp} \mod \pi \latticeL{k}
\]

obtained in Proposition <ref>. Applying the operator $\sum\limits_{a = 0}^{p - 1}a^{n - \frac{r - 1}{2}}\begin{pmatrix}1 & 0 \\ -a & 1\end{pmatrix}$ to both sides,

substituting $\lambda = a + b$ on the right side and summing over $b$ using
equation (<ref>) with $i = \dfrac{r+1}{2}$,

replacing $\lambda$ by $a$, and substituting the value of $c(n, j)$ obtained in
Proposition <ref>,

we get

\begin{eqnarray}\label{Applied inductive steps for g1 in self-dual}
0 \equiv \sum_{a = 0}^{p - 1}\sum_{j = r - n}^{\frac{r - 1}{2}}{n - \frac{r - 1}{2} \choose n - r + j}\frac{(r - n)!n}{(r - j) \cdots (n - j)}a^{\frac{r + 1}{2} - j}(z - a)^j\mathbbm{1}_{a + p\Zp} \mod \pi \latticeL{k}.
\end{eqnarray}

Next, consider the following equation obtained from the binomial expansion

\begin{eqnarray}\label{Trivial equation in g1 in self-dual}