This last property will be used to control the quantities $\gamma_1$ and $\gamma_2$
appearing in the right-hand sides of the estimates of Theorem <ref>.

From these estimates, and recalling that $R=1$ and $\omega_0=1$,
the corollary follows immediately after realizing that $\|(f+g)_-\|_{L^\infty(B_1)}\le\|u\|_{L^\infty(B_1)}$.
Indeed, since $u-g\le f$ everywhere in $B_1$, we have $-(f+g)\le-u\le\|u\|_{L^\infty(B_1)}$,
from which the previous bound follows.

§ GENERAL INTEGRO-DIFFERENTIAL OPERATORS

§.§ Equivalent formulations of
the key inequalities

In this section we
provide the proof of Proposition <ref>. The proposition will be used in next subsection to provide the main step towards first derivative estimates.
The proof of Proposition <ref>
will also establish the following result, a variant
of it which involves the positive part of the derivative. If one could prove  that inequality (<ref>) appearing below holds for an appropriate error $E$, then one-sided second derivative estimates would follow;  see Open problem <ref>.

Let ${\mathcal{L}}_K$ be
as in (<ref>), with $K$ satisfying (<ref>)
and (<ref>).
Given $v\in C^\infty(\R^n)\cap W^{1,\infty}(\R^n)$,
$\overline\eta \in
C^\infty(\R^n)\cap L^\infty(\R^n)$,
$e\in \R^n$ with $|e|=1$, $\tau>0$, and $\CALERRE\in\R$,

\begin{equation}\label{8uj8ij6yhb6ytg8uj} \psi:=\overline\eta^2 \big(
\partial_e v\big)^2_{+} +\tau v^2.\end{equation}

Then, the inequality

\begin{equation}\label{MAY18-3}
{\mathcal{L}}_K \psi\le 2\overline\eta^2\,(\partial_e v)_{+}\,
{\mathcal{L}}_K\big(( \partial_e v)_{+}\big)
+2\tau v\,{\mathcal{L}}_K v+\CALERRE
\end{equation}

holds at a point $x\in\R^n$ if and only if

\begin{equation}\label{MAY18-4}\begin{split}&
2\int_{\R^n}\overline\eta(x)\,\big( \overline\eta(x)-\overline\eta(y)\big)\,(\partial_e v)_{+}(x)\,(\partial_e v)_{+}(y)\,K(x-y)\,dy
\\ &\qquad\le\,\int_{\R^n}\big|\overline\eta(x)\,(\partial_e v)_{+}(x)-\overline\eta(y)\,(\partial_e v)_{+}(y)\big|^2\,K(x-y)\,dy
\\&\qquad\qquad+\tau\int_{\R^n}\big|v(x)-v(y)\big|^2\,K(x-y)\,dy

Note that, as in previous sections, ${\mathcal{L}}_K \psi$ is well defined everywhere in $\R^n$
since $(\partial_e v)_{+}^2$ is a locally $C^{1,1}$ function which is bounded in all of $\R^n$.
However, $(\partial_e v)_{+}$ in the right-hand side
of (<ref>)
is only a $C^{1,1}$ function by below (locally). Recall that one says that $\varphi\in C(B_R)$ is “$C^{1,1}$
from below” in $B_R$ if for every $x_0\in B_R$ there exists $w\in C^{1,1}(B_R)$ such that $w\le \varphi$ everywhere in $B_R$ and $w(x_0)=\varphi(x_0)$.
This setting is sufficient to define the operator pointwise everywhere by having values
in $\{-\infty\}\cup\R$, with the convention
that $
0\cdot(-\infty)=0$ in the expression $(\partial_e v)_{+}\,
{\mathcal{L}}_K\big(( \partial_e v)_{+}\big)$.

Furthermore, we notice that whenever (<ref>) holds true, then also

\begin{equation}\label{chiamare}
{\mathcal{L}}_K \psi\le 2\overline\eta^2\,(\partial_e v)_{+}\,
{\mathcal{L}}_K  \partial_e v
+2\tau v\,{\mathcal{L}}_K v+\CALERRE,
\end{equation}

since this would follow from the above convention when $\partial_e v(x)\le0$
and from the fact that

$${\mathcal{L}}_K\big(( \partial_e v)_{+}\big)(x)\le{\mathcal{L}}_K\partial_e v(x)\;\mbox{ when }\;\partial_e v(x)>0.$$

This observation
is relevant since
Lemma <ref> gives control of ${\mathcal{L}}_K\partial_e v$ by above.
This is why, within Section <ref> on operators with an extension, the inequality was stated
as (<ref>) (with $E=0$), and not as (<ref>).

We also point out that the integrals in (<ref>)
and (<ref>) are all
well defined, due to the regularity of
the functions involved. First, the integral in the left-hand side of (<ref>)
is well defined in the principal value sense, since, for small $z$,

\begin{eqnarray*}&&
\big( \eta(x)-\eta(x+z)\big)\,\partial_e u(x+z)\,K(z)=
\big( -\nabla\eta(x)\cdot z+O(|z|^2)\big)\,
\big(\partial_e u(x)+O(|z|)
\big)\,K(z)\\&&\qquad=-\partial_e u(x)\nabla\eta(x)\cdot z\,K(z)
\end{eqnarray*}

and the term $-\nabla\eta(x)\cdot z\,K(z)$
provides a null contribution to the principal value of the integral
over $z\in B_1$,
thanks to the symmetry assumption (<ref>)
(the decay assumption (<ref>)
will then make the term $O(|z|^2)\,K(z)$ integrable for $z\in B_1$).
A similar argument applies to the first integral in (<ref>) where one may assume $(\partial_e v)_+(x)>0$.

The second integral in (<ref>) is instead a classical
Lebesgue integral,
since, for small $z$,

$$ \big|\eta(x)\,\partial_e u(x)-\eta(x+z)\,\partial_e u(x+z)\big|^2\,K(z)
\le \|\eta\,\partial_e u\|_{W^{1,\infty}(B_1(x))}^2\,|z|^2\,K(z).$$

The same argument applies to
the first integral in the right-hand side of (<ref>), since $(\partial_e v)_+$ is locally a $W^{1,\infty}$ function.
A simpler argument gives that also the last integrals in (<ref>)
and (<ref>) are well defined.

We now establish both Propositions <ref>
and <ref> in a unified manner.

Here, we take ${\mathcal{G}}(t)$ to be[We point out that
formally (that is, when all the integrals make sense) the arguments
presented here are valid for all nonlinear functions ${\mathcal{G}}$.]
either $t$ or $t_+$. We also adopt the notation ${\mathcal{G}}^2(t):=\big({\mathcal{G}}(t)\big)^2$. We will do the computation with $u$, $\sigma$, and $\eta$
to address Proposition <ref> (with ${\mathcal{G}}(t)=t$ in this case),
and we replace these choices by $v$, $\tau$, and $\overline\eta$, respectively,
to complete the proof of Proposition <ref> (with ${\mathcal{G}}(t)=t_+$ in this case).

Given a kernel $K$, we have

\begin{eqnarray*}
&& {\mathcal{L}}_K \Big( \eta^2\,{\mathcal{G}}^2\big(\partial_e u\big)+\sigma
\,u^2\Big)(x)
-2\eta^2(x)\, {\mathcal{G}}\big(\partial_e u(x)\big)\,{\mathcal{L}}_K\,{\mathcal{G}}\big(\partial_e u\big)(x)
\\&&\qquad\qquad-2\sigma\,u(x)\,{\mathcal{L}}_K u(x)+\sigma\int_{\R^n}\big|u(x)-u(y)\big|^2\,K(x-y)\,dy\\
&&\qquad= \int_{\R^n}
\Big( \eta^2(x)\,{\mathcal{G}}^2\big(\partial_e u(x)\big)
-\eta^2(y)\,{\mathcal{G}}^2\big(\partial_e u(y)\big)\Big)\,K(x-y)\,dy\\&&\qquad\qquad
+\sigma\int_{\R^n} \big(u^2(x)-u^2(y)\big)\,K(x-y)\,dy\\
&&\qquad\qquad-2\eta^2(x)\, {\mathcal{G}}\big(\partial_e u(x)\big)\,
\int_{\R^n}\Big( {\mathcal{G}}\big(\partial_e u(x)\big)-{\mathcal{G}}\big(\partial_e u(y)\big)\Big)
\,K(x-y)\,dy\\
&&\qquad\qquad-2\sigma\,u(x)\,\int_{\R^n} \big(u(x)-u(y)\big)\,K(x-y)\,dy
\\&&\qquad= \int_{\R^n}
\Big( \eta^2(x)\,{\mathcal{G}}^2\big(\partial_e u(x)\big)
-\eta^2(y)\,{\mathcal{G}}^2\big(\partial_e u(y)\big)\Big)\,K(x-y)\,dy
\\&&\qquad\qquad-2\eta^2(x)\, {\mathcal{G}}\big(\partial_e u(x)\big)\,
\int_{\R^n}\Big( {\mathcal{G}}\big(\partial_e u(x)\big)-{\mathcal{G}}\big(\partial_e u(y)\big)\Big)
\,K(x-y)\,dy\\
&&\qquad= \int_{\R^n}\Big( 2\eta^2(x)\, {\mathcal{G}}\big(\partial_e u(x)\big)\, {\mathcal{G}}\big(\partial_e u(y)\big)
\\&&\qquad\qquad-\eta^2(x)\, {\mathcal{G}}^2\big(\partial_e u(x)\big)-
\eta^2(y)\, {\mathcal{G}}^2\big(\partial_e u(y)\big)
\Big)\,K(x-y)\,dy
\\ &&\qquad=2\int_{\R^n} \eta(x)\,\big(\eta(x)-\eta(y)\big)\, {\mathcal{G}}\big(\partial_e u(x)\big)\, {\mathcal{G}}\big(\partial_e u(y)\big)
\,K(x-y)\,dy
\\&&\qquad\qquad-
\int_{\R^n}\Big|
\eta(x)\, {\mathcal{G}}\big(\partial_e u(x)\big)-
\eta(y)\, {\mathcal{G}}\big(\partial_e u(y)\big)
\Big|^2\,K(x-y)\,dy.
\end{eqnarray*}

As a consequence, the inequality

$$ {\mathcal{L}}_K \big(\eta^2
{\mathcal{G}}^2 (\partial_e u )
+\sigma u^2\big)\le 2\eta^2\,{\mathcal{G}}\big(\partial_e u\big)\,{\mathcal{L}}_K\,
{\mathcal{G}}\big(\partial_e u\big) +2\sigma u\,{\mathcal{L}}_K u+\CALERRE

is pointwise equivalent to the inequality

\begin{equation*}
\begin{split}&
2\int_{\R^n}\eta(x)\,\big( \eta(x)-\eta(y)\big)\,{\mathcal{G}}\big(
\partial_e u(x)\big)\,{\mathcal{G}}\big(\partial_e u(y)\big)\,K(x-y)\,dy
\\ &\qquad\le\,\int_{\R^n}\Big|\eta(x)\,{\mathcal{G}}\big(\partial_e u(x)\big)
-\eta(y)\,{\mathcal{G}}\big(\partial_e u(y)\big)\Big|^2\,K(x-y)\,dy
\\&\qquad\qquad+\sigma\int_{\R^n}\big|u(x)- u(y)\big|^2\,K(x-y)\,dy

From this, one deduces Propositions <ref> and <ref>.

§.§ Proof of the first key inequality with a remainder

This subsection contains the proof
of our main inequality for the auxiliary function in the case of general
integro-differential operators.

By the translation invariance of the problem, we see that, to establish (<ref>)
in $B_2$,
it suffices to prove that

\begin{equation}\label{T6.3}\begin{split}&
\eta(0)-\eta(y)\big)\partial_e u(0)\,\partial_e u(y)\,K(y)\,dy\\ &\qquad\qquad\leq\,
\int_{\R^n} \big|
\eta(0)\partial_e u(0)-\eta(y)\partial_e u(y)\big|^2\,K(y)\,dy\\&\qquad\qquad\qquad+
\sigma_\e\,\int_{\R^n}\big|u(0)-u(y)\big|^2\,K(y)\,dy+\e^2\,\| \partial_e u\|^2_{L^\infty(B_1)},
\end{split}
\end{equation}

where we have used
Proposition <ref> at $x=0$
with $\CALERRE:=\e^2\,\|\partial_e u\|^2_{L^\infty(B_1)}$.
In this way,
the right-hand side of inequality (<ref>)
in $B_2$ becomes $\e^2\,\| \partial_e u\|^2_{L^\infty(B_3)}$.

To prove (<ref>), we exploit an appropriate cutoff procedure
on the gradient of $u$, to suitably remove the singularity
of the integrand near the origin in the left-hand side of (<ref>), (without spoiling the estimates
at infinity).
we consider an odd function $\xi\in C^\infty_c((-2,2))$
such that $\xi(t)=t$ if $t\in(-1,1)$, and
$|\xi(t)|\le2$ for every $t\in\R$.
We also consider $\delta\in\left(0,\frac12\right]$, and set $\xi_\delta(t):=\delta\xi(t/\delta)$.
We observe that

\begin{equation}\label{QUI90}
\xi_\delta(0)=0,\quad \xi_\delta'(0)=1, \quad{\mbox{ and }}\quad
\|\xi_\delta\|_{C^2(\R)}\le \frac{C}{\delta},
\end{equation}

for some universal constant $C$.

We also introduce the map $
\R^n\ni y=(y_1,\dots,y_n)\mapsto {\mathcal{Z}}_\delta(y):=\big( \xi_\delta(y_1),\dots,\xi_\delta(y_n)\big)$.
Since $\xi_\delta$ is odd, we have that

\begin{equation}\label{QUI2091} \int_{\R^n} \nabla \eta(0)\cdot {\mathcal{Z}}_\delta(y)\,K(y)\,dy

in the principal value sense, thanks to the symmetry of the kernel (<ref>).
Consequently, we have that

\begin{equation}\label{QUI91}\begin{split}&
\int_{\R^n}\eta(0)\,\big(
\eta(0)-\eta(y)\big)\partial_e u(0)\,\partial_e u(y)\,K(y)\,dy\\ &\qquad\qquad=\,
\int_{\R^n}\Big(\eta(0)\,\big(
\eta(0)-\eta(y)
\big)\partial_e u(0)\,\partial_e u(y)+\eta(0)|\partial_e u(0)|^2\,
\nabla\eta(0)\cdot {\mathcal{Z}}_\delta(y)\Big)
\,K(y)\,dy.
\end{split}\end{equation}

Furthermore, we set

\begin{equation}\label{QUI92}
I_1(y)\, :=\, \eta(0)\,\Big(\eta(0)-\eta(y)
+\nabla\eta(0)\cdot {\mathcal{Z}}_\delta(y)
\Big)\partial_e u(0)\,\partial_e u(y),
\end{equation}

\begin{equation}\label{QUI92BIS}
I_2 (y)\,:=\,
\partial_e u(0)\,\big(\eta(0)\partial_e u(0)-\eta(y)\partial_e u(y)\big)\,
\nabla\eta(0)\cdot {\mathcal{Z}}_\delta(y),\end{equation}

\begin{equation}\label{I3}
\big(\eta(y)-\eta(0)\big)\,
\partial_e u(0)\,\partial_e u(y)\,
\nabla\eta(0)\cdot {\mathcal{Z}}_\delta(y),
\end{equation}

and we
point out that

\begin{equation}\label{0x-10133}\begin{split}
& \eta(0)\,\big(\eta(0)-\eta(y)\big)\partial_e u(0)\,\partial_e u(y)
+\eta(0)|\partial_e u(0)|^2\,
\nabla\eta(0)\cdot {\mathcal{Z}}_\delta(y)\\ &\qquad\qquad
\eta(0)\,\Big(\eta(0)-\eta(y)
+\nabla\eta(0)\cdot {\mathcal{Z}}_\delta(y)
\Big)\partial_e u(0)\,\partial_e u(y)\\
&\qquad\qquad\qquad+\eta(0)\partial_e u(0)\,\big(\partial_e u(0)-\partial_e u(y)\big)\,
\nabla\eta(0)\cdot {\mathcal{Z}}_\delta(y)\\&\qquad\qquad
=\; I_1(y)+
\partial_e u(0)\,\big(\eta(0)\partial_e u(0)-\eta(y)\partial_e u(y)\big)\,
\nabla\eta(0)\cdot {\mathcal{Z}}_\delta(y)\\&\qquad\qquad\qquad
+\partial_e u(0)\,\big(\eta(y)\partial_e u(y)-\eta(0)\partial_e u(y)\big)\,
\nabla\eta(0)\cdot {\mathcal{Z}}_\delta(y)
\\ &\qquad\qquad=\; I_1(y)+I_2(y)+I_3(y).
\end{split}\end{equation}

Hence, substituting into (<ref>), we obtain

\begin{equation}\label{QUI220}
\int_{\R^n}\eta(0)\,\big(
\eta(0)-\eta(y)\big)\partial_e u(0)\,\partial_e u(y)\,K(y)\,dy =
\int_{\R^n} \big( I_1(y)+I_2(y)+I_3(y)\big)\,K(y)\,dy.
\end{equation}

Now, we set

\begin{equation}\label{QUI99} \varphi_\delta(y):=\eta(0)-\eta(y)
+\nabla\eta(0)\cdot {\mathcal{Z}}_\delta(y),\end{equation}

and we observe that

\begin{equation}\label{QUI100}
\varphi_\delta(0)=0,\quad\nabla\varphi_\delta(0)=0,\quad{\mbox{ and }}\quad
\|\varphi_\delta\|_{C^2(\R^n)}\le C_\delta\,\|\eta\|_{C^2(\R^n)},
\end{equation}

for some constant $C_\delta>0$, depending only on $n$ and $\delta$,
thanks to the properties (<ref>) of $\xi_\delta$.

We also set

\begin{equation}\label{QUI101}
{ J_1 }\, :=\,\int_{\R^n}\partial_e\big(\varphi_\delta(y) K(y)\big)\,\big(
\eta(y)\,\partial_e u(y)-\eta(0)\,\partial_e u(0) \big)\big(u(y)-u(0)\big)\,dy\end{equation}

\begin{equation}\label{QUI101E}
{ J_2 }\,:=\,
\frac12\,
\int_{\R^n}\partial_e\,\Big(\partial_e \big(\varphi_\delta(y) K(y)\big)\eta(y)\Big)
\,\big|u(y)-u(0)\big|^2\,dy.
\end{equation}

We now perform some integration by parts in $B_R$ and use that the boundary terms
on $\partial B_R$ go to zero as $R\to+\infty$ (as well as that the integrands are integrable in $\R^n$),
thanks to
the decay of the kernel and of its derivatives assumed in (<ref>) and (<ref>).
from (<ref>) and (<ref>), and integrating by parts
we find that

\begin{equation}\label{QUI107}
\begin{split}
\int_{\R^n} I_1(y)\,K(y)\,dy\,&
=\,\int_{\R^n}\eta(0)\,\varphi_\delta(y)\,\partial_e u(0)\,\partial_e u(y)\,K(y)\,dy\\
&=\,\int_{\R^n}\varphi_\delta(y)\,\eta(0)\,\partial_e u(0)\,\partial_e \big(u(y)-u(0)\big)\,K(y)\,dy
\\&=\,-
\int_{\R^n}\partial_e \big(\varphi_\delta(y) K(y)\big)\,\eta(0)\,\partial_e u(0) \big(u(y)-u(0)\big)\,dy
\\ &=\,
\int_{\R^n}\partial_e \big(\varphi_\delta(y) K(y)\big)\big(
\eta(y)\,\partial_e u(y)-\eta(0)\,\partial_e u(0) \big)\big(u(y)-u(0)\big)\,dy
\\&\qquad\qquad-
\int_{\R^n}\partial_e\big(\varphi_\delta(y) K(y)\big)\eta(y)\,\partial_e u(y) \big(u(y)-u(0)\big)\,dy
\\&
{ J_1 }-\frac12\,
\int_{\R^n}\partial_e \big(\varphi_\delta(y) K(y)\big)\,\eta(y)\,\partial_e
\big|u(y)-u(0)\big|^2\,dy
\\&
{ J_1 }+\frac12\,
\int_{\R^n}{\partial_e}\,\Big( \partial_e \big(\varphi_\delta(y) K(y)\big)\eta(y)\Big)
\,\big|u(y)-u(0)\big|^2\,dy\\&
=\, { J_1 }+{ J_2 }.
\end{split}
\end{equation}

Furthermore, recalling again
the bound (<ref>) on the first and second derivatives
of the kernel and (<ref>),
we point out that

\begin{equation}\label{QUI103}
\begin{split}&
\Big| \partial_e \big(\varphi_\delta(y) K(y)\big) \Big|\le
\big| \partial_e \varphi_\delta(y) \big|\,K(y)+\big|\varphi_\delta(y)\big|\,\big|
\partial_e K(y)\big|\\ &\qquad\le
C_{\delta,\eta} \big( K(y)+|y|\,\big|\partial_e K(y)\big|\big)
\le C_{\delta,\eta}\,K(y),\end{split}
\end{equation}

for some constant $C_{\delta,\eta}>0$, possibly varying from line to line,
and depending
only on $n$, $\delta$,
and on the structural constant $\CUNO$
in (<ref>)
(for convenience, in what follows, we will rename $C_{\delta,\eta}$
allowing dependences also
on $s$ and on the constants $\CDUEDUE$ and $\CDUE$ in (<ref>)).

In addition, using again (<ref>) and (<ref>),

\begin{eqnarray*}
\Big|\partial^2_e\big(\varphi_\delta(y) K(y)\big)\eta(y)\Big|&\le&
\Big|\partial_e^2\varphi_\delta(y) K(y)\Big|+
\Big|\partial_e \varphi_\delta(y)\,\partial_e K(y)\Big|+
\Big|\varphi_\delta(y) \,\partial_e^2 K(y)\Big|
\right)\\
&\le& C_{\delta,\eta}\,
\Big(K(y)+|y|\,\big|\partial_e K(y)\big|+|y|^2\big|D^2 K(y)\big|
\Big)\\
&\le& C_{\delta,\eta}\,K(y).
\end{eqnarray*}

Hence, from the latter estimate and (<ref>),

\begin{equation}\label{QUI105}
\Big|\partial_e \,\Big(\partial_e \big(\varphi_\delta(y) K(y)\big)\eta(y)\Big)\Big|
\le C_{\delta,\eta}\,K(y).
\end{equation}

Thus, by the definitions (<ref>) and (<ref>)
of $J_1$ and $J_2$, and the estimates (<ref>) and (<ref>),
using an appropriate Cauchy-Schwarz inequality
we get that

\begin{equation}\label{VECCHIA J1J2}\begin{split}&
|{ J_1 }|+|{ J_2 }|\\
&\qquad\le\, C_{\delta,\eta}\,\left(
\int_{\R^n}
\big|
\eta(y)\,\partial_e u(y)-\eta(0)\,\partial_e u(0) \big|\,\big|u(y)-u(0)\big|\,K(y)\,dy\right.\\&\qquad\qquad\qquad\left.+
\int_{\R^n} \big|u(y)-u(0)\big|^2\,K(y)\,dy\right)\\&\qquad
\le\,\frac18\,
\int_{\R^n}
\big|\eta(y)\,\partial_e u(y)-\eta(0)\,\partial_e u(0) \big|^2\,K(y)\,dy+C_{\delta,\eta}\,
\int_{\R^n} \big|u(y)-u(0)\big|^2\,K(y)\,dy.
\end{split}\end{equation}

This and (<ref>) give that

\begin{equation}\label{QUI114}\begin{split}&
\int_{\R^n} I_1(y)\,K(y)\,dy\\&\qquad
\le\;
\frac18\,
\int_{\R^n}
\big|\eta(y)\,\partial_e u(y)-\eta(0)\,\partial_e u(0) \big|^2\,K(y)\,dy+C_{\delta,\eta}\,
\int_{\R^n} \big|u(y)-u(0)\big|^2\,K(y)\,dy.
\end{split}\end{equation}

Next, we notice that $|{\mathcal{Z}}_\delta(y)|\le C\,|y|$, for every $y\in\R^n$, for some constant $C$ depending only
on $n$.
Thus, we have that

\begin{equation} \label{QUI109} J_3:=
\int_{B_1} |y|\, |{\mathcal{Z}}_\delta(y)|\,K(y)\,dy<+\infty.\end{equation}

In view of the definition (<ref>) of $I_2$,
noticing that ${\mathcal{Z}}_\delta$ is supported in $B_1$,
and using a suitable Cauchy-Schwarz inequality,
we have that

\begin{equation}\label{QUI111}\begin{split}
\int_{\R^n} I_2(y)\,K(y)\,dy\\ &\qquad\le\,C_{\eta}\,
\int_{B_1} \big| \partial_e u(0)\big|
\big|\eta(0)\partial_e u(0)-\eta(y)\partial_e u(y)\big|\;|{\mathcal{Z}}_\delta(y)|\,K(y)\,dy\\&\qquad
\le\,
\frac18\,
\int_{\R^n}
\big|\eta(0)\partial_e u(0)-\eta(y)\partial_e u(y)\big|^2\,K(y)\,dy
\int_{B_1} \big| \partial_e u(0)\big|^2\;|{\mathcal{Z}}_\delta(y)|^2\,K(y)\,dy\\&\qquad
\le\, \frac18\,
\int_{\R^n}
\big|\eta(0)\partial_e u(0)-\eta(y)\partial_e u(y)\big|^2\,K(y)\,dy
+C_{\eta}\,\|\partial_e u\|_{L^\infty(B_1)}^2\,J_3
\end{split}\end{equation}

for some constant $C_\eta>0$ depending only
on $n$ and $\| \eta\|_{C^2(\R^n)}$.
Similarly, recalling the definition (<ref>)
of $I_3$,

\begin{equation}\label{QUI112}\begin{split}
\int_{\R^n} I_3(y)\,K(y)\,dy\,\le\,&C_{\eta}\,
\int_{B_1}\big|\eta(y)-\eta(0)\big|\,
|\partial_e u(0)|\,|\partial_e u(y)|\;|
\le\,&C_{\eta}\,
\int_{B_1}|y|\,
|\partial_e u(0)|\,|\partial_e u(y)|\;|
\le\,&C_{\eta}\,\|\partial_e u\|_{L^\infty(B_1)}^2\,J_3.
\end{split}\end{equation}

Now we provide a bound on $J_3$. To this aim, we split
the integral computation inside $B_\delta$, where ${\mathcal{Z}}_\delta(y)=y$,
and in $B_1\setminus B_\delta$, where $|{\mathcal{Z}}_\delta(y)|\le 2\delta\,n^{1/2}$.
In this way, recalling the definition (<ref>)
of $J_3$, we get that

\begin{eqnarray*} |J_3|
&\le& C\,\left(
\int_{B_\delta} |y|^2\,K(y)\,dy
+\delta\int_{B_1\setminus B_\delta} |y|\, K(y)\,dy
\right),
\end{eqnarray*}

for some constant $C$ depending only on $n$.
The latter quantity tends to zero as $\delta\searrow0$,
thanks to the bounds (<ref>) on the kernel.
Given $\e>0$, we can therefore
take $\delta\in(0,1)$ sufficiently small,
in such a way that

\begin{equation*} |J_3|\le\e^2.\end{equation*}

After this choice of $\delta$,
the constants $C_{\delta,\eta}$ above
will be written accordingly,
with a slight abuse of notation, as $C_{\e,\eta}$.
As a consequence, collecting the estimates in (<ref>),
(<ref>), and (<ref>), we conclude that

\begin{eqnarray*}&&
\int_{\R^n} \big(I_1(y)+I_2(y)+I_3(y)\big)\,K(y)\,dy\\&&\qquad\qquad
\le \frac14\,
\int_{\R^n}
\big|\eta(y)\,\partial_e u(y)-\eta(0)\,\partial_e u(0) \big|^2\,K(y)\,dy\\
\int_{\R^n} \big|u(y)-u(0)\big|^2\,K(y)\,dy+
C_{\eta}\,\e^2\,\|\partial_e u\|_{L^\infty(B_1)}^2.
\end{eqnarray*}

Therefore, recalling (<ref>), we deduce that

\begin{eqnarray*}&&
\int_{\R^n}\eta(0)\,\big(
\eta(0)-\eta(y)\big)\partial_e u(0)\,\partial_e u(y)\,K(y)\,dy
\\&&\qquad\qquad
\le \frac14\,
\int_{\R^n}
\big|\eta(y)\,\partial_e u(y)-\eta(0)\,\partial_e u(0) \big|^2\,K(y)\,dy\\
\int_{\R^n} \big|u(y)-u(0)\big|^2\,K(y)\,dy+
C_{\eta}\,\e^2\,\|\partial_e u\|_{L^\infty(B_1)}^2.
\end{eqnarray*}

This establishes (<ref>), up to renaming $\e$ and the constants,
by choosing $\sigma_\e$ large enough.
As a consequence, also (<ref>) follows.

§.§ Proof of Theorem <ref>

First of all, we show that the structure of
equation (<ref>) is preserved under rescaling. For this,
given $R>0$, we set

\begin{equation}\label{912ou435uuuquq} F_R(p):=R^{2s}\,F(R^{-2s}p).\end{equation}

We observe that

\begin{equation}\label{TRA5}
\begin{split}&
{\mbox{$F_R$ satisfies the same structural
assumptions as~$F$ given in~\eqref{CONV:POSITIVA}--\eqref{CONV:F},}}\\&
{\mbox{with the same constant~$\vartheta_0$.}}\end{split}\end{equation}

For every $s\in(0,1)$ and $ R >0$, given a kernel $K$
satisfying (<ref>), (<ref>), and (<ref>),
we also consider the rescaled kernel $ K^{[ R ]}(z)$ defined
by $
K^{[ R]}(z):=
Notice that

\begin{equation}\begin{split}\label{STRA5}&
{\mbox{conditions~\eqref{EVEN}, \eqref{KLIM}, and~\eqref{KC1} are automatically satisfied by~$K^{[ R ]}$,}}
\\&{\mbox{with the same
constants~$\CDUEDUE$, $\CDUE$,
independently of~$R$.}}\end{split}\end{equation}

Hence, a direct computation shows that, for every $j\in\{1,\dots,J\}$,
if $u_R(x):=u(Rx)$,

$$ {\mathcal{L}}_{K_j} u(Rx)= R^{-2s}\,{\mathcal{L}}_{K_j^{[ R ]}} u_R(x).$$

From this and (<ref>) we have that, if $u
\in C^\infty(B_R)\cap W^{1,\infty}(\R^n)$ is a
solution of (<ref>)
everywhere in $B_R$,
then, $u_R$ is a solution of

\begin{equation}\label{TRA} F_R\left( {\mathcal{L}}_{K_1^{[R ]}}u_R-g_1^{[R]}, \;\cdots,\; {\mathcal{L}}_{K_J^{[ R ]}}u_R-g_J^{[R]}\right)=0\quad{\mbox{everywhere in }}B_1,\end{equation}

where $g_j^{[R]}(x):=R^{2s}g_j(Rx)$.

Now, we exploit this and Theorem <ref>
to establish the following scaled first derivative estimate. From it, we will see that
Theorem <ref> follows (up to scaling
the radius of the ball).

For $\rho>0$, assume that $u
\in C^\infty(\R^n)\cap W^{1,\infty}(\R^n)$ is a
solution of (<ref>)
in $B_{2\rho}$ (instead of $B_R$), under the structural
assumptions in (<ref>), (<ref>), and (<ref>).

Then, for every $\e>0$ there exists $C_\e>0$, depending only
on $\e$, $n$, $\vartheta_0$,
on the structural constants $s$, $\CDUEDUE$,
$\CDUE$, and $\CUNO$
in (<ref>) and (<ref>),
such that

\begin{eqnarray*}
\rho\,\|\partial_e u\|_{L^\infty(B_\rho)}&\le&
C_\e\,\left( \|u\|_{L^\infty(\R^n)}
+\rho^{1+2s}\gamma_1+\big( \rho^{2s}
\,\|(f+g)_-\|_{L^\infty(B_{2\rho})}\,
\|u\|_{L^\infty(B_{2\rho})}\big)^{1/2}\right)\\&&\qquad+
\e\rho\,\|\partial_e u\|_{L^\infty(B_{3\rho})},\end{eqnarray*}

for every $e\in\R^n$ with $|e|=1$. Here above, $\gamma_1$
is the constant defined right before Theorem <ref>
(here with $R$ replaced by $2\rho$)
and $g$ is as in (<ref>).

Thanks to (<ref>), (<ref>),
and (<ref>) (applied with $R:=2\rho$),
up to scaling, we may assume that $\rho:=1$.
In this way, the equation is supposed to hold true in $B_2$.

Let $\eta$ and $\sigma_\e$ be as in Theorem <ref>, and

$$ \Phi:=\eta^2|\partial_e u|^2+\sigma_\e \left(u-\sup_{B_2}u\right)^2

We consider the operator $L$ introduced in (<ref>)
and notice that, by Lemma <ref>
and Theorem <ref>, we have, everywhere
in $B_2$,

\begin{eqnarray*}
L \Phi&\le&
2\eta^2\partial_e u \,L\partial_e u+
2\sigma_\e \,\left(u-\sup_{B_2}u\right)\,L u+\e^2\,\|\partial_e u\|^2_{L^\infty(B_3)}
\\ &\le&
2\eta^2|\partial_e u| \left(
\|\partial_e f\|_{L^\infty(B_2)}+\sum_{j=1}^J\alpha_j\|\partial_e g_j\|_{L^\infty(B_2)} \right)-
2\sigma_\e \,\left(\sup_{B_2}u-u\right)\,(f+g)\\&&\qquad
+\e^2\,\|\partial_e u\|^2_{L^\infty(B_3)}
\\&\le&
\gamma_0,
\end{eqnarray*}

\begin{equation}\label{CGHUIXTuhcjiYA}\begin{split}
\gamma_0:=\,&
2\sup_{B_2}\left(\eta^2|\partial_e u| \left(\|\partial_e f\|_{L^\infty(B_2)}+\sum_{j=1}^J\alpha_j\|
\partial_e g_j\|_{L^\infty(B_2)}
\right)\right)\\&\qquad+
4\sigma_\e \,\|u\|_{L^\infty(B_2)}\,\| (f+g)_-\|_{L^\infty(B_2)}
+\e^2\,\|\partial_e u\|^2_{L^\infty(B_3)}

Consequently, by the maximum principle in Proposition <ref>
(used here with $R:=2$),

\begin{equation*}
\begin{split}&
\sup_{B_2}\left(\eta^2|\partial_e u|^2+\sigma_\e \left(u-\sup_{B_2}u\right)^2\right)
\sup_{B_2}\Phi\le\sup_{\R^n\setminus B_2}\Phi
\le
\sup_{\R^n\setminus B_2}\left(\eta^2|\partial_e u|^2+\sigma_\e \left(
\end{split}
\end{equation*}

for some constant $C$, depending only on $n$, $\vartheta_0$, and $C_1$,
and possibly varying from step to step of this proof.

Taking now $\eta$ such that $0\le\eta\le1$, $\eta:=1$ in $B_1$,
and $\eta:=0$ outside $B_2$, recalling (<ref>),
and using the Cauchy-Schwarz inequality,
we conclude that

\begin{eqnarray*}
\sup_{B_2}(\eta^2 |\partial_e u|^2)&\le&4 \sigma_\e\,\|u\|^2_{L^\infty(\R^n)}+
C \gamma_0
\\&\le&4 \sigma_\e\,\|u\|_{L^\infty(\R^n)}^2+
C\sup_{B_2}\left(\eta^2|\partial_e u| \left(\|\partial_ef\|_{L^\infty(B_2)}+\sum_{j=1}^J\alpha_j
\|\partial_e g_j\|_{L^\infty(B_2)}
\right)
\right)\\&&\qquad+
C\sigma_\e \,\|(f+g)_-\|_{L^\infty(B_2)}\,\|u\|_{L^\infty(B_2)}+
C\e^2\,\|\partial_e u\|_{L^\infty(B_3)}^2\\
\sigma_\e\,\|u\|_{L^\infty(\R^n)}^2+\frac12\,\sup_{B_2}(\eta^2|\partial_e u|^2)
+C\|\partial_ef\|_{L^\infty(B_2)}^2+C \sum_{j=1}^J\alpha_j^2 \|\partial_e g_j\|^2_{L^\infty(B_2)}
\\&&\qquad+
C\sigma_\e \,\|(f+g)_-\|_{L^\infty(B_2)}\,\|u\|_{L^\infty(B_2)}+
C\e^2\,\|\partial_e u\|_{L^\infty(B_3)}^2.
\end{eqnarray*}

Consequently, reabsorbing the second term on the left-hand side,

\begin{eqnarray*}
\sup_{B_1}|\partial_eu|^2&\le&
\sup_{B_2}(\eta^2 |\partial_e u|^2)
\\ &\le& 8
\sigma_\e\,\|u\|_{L^\infty(\R^n)}^2+C\|\partial_ef\|_{L^\infty(B_2)}^2
+C \,\sum_{j=1}^J\alpha_j^2 \|\partial_e g_j\|^2_{L^\infty(B_2)}
\\&&\quad+
C\sigma_\e \,\|(f+g)_-\|_{L^\infty(B_2)}\,\|u\|_{L^\infty(B_2)}+
C\e^2\,\|\partial_e u\|_{L^\infty(B_3)}^2,
\end{eqnarray*}

which yields the desired result, up to renaming constants
(including $\epsilon$ so that $C\epsilon^2$ becomes $\epsilon^2$ as in the statement;
note that $C$ is independent of $\epsilon$).

Theorem <ref>
now follows from
Proposition <ref> and
Lemma <ref>. The details go as follows.

Up to scaling, it suffices to prove the desired result for $R:=4$.
For this, we exploit Proposition <ref>
in every ball $B_{2\rho}(x)$,
for all $x\in B_2$
and $\rho\in(0,1)$, and conclude that

\begin{equation*}\begin{split}&\rho
\|\partial_e u\|_{L^\infty(B_{\rho}(x))}\le
+\rho^{1+2s}\gamma_1+ \rho^{s}
\,\|(f+g)_-\|_{L^\infty(B_{2\rho}(x))}^{1/2}\,
\|u\|_{L^\infty(B_{2\rho}(x))}^{1/2}\right)\\&\qquad\qquad\qquad\qquad\qquad+
\e \rho\,\|\partial_e u\|_{L^\infty(B_{3\rho}(x))},\end{split}
\end{equation*}

with $\gamma_1$
as stated in Theorem <ref> (with $R:=4$, as mentioned above).

In particular, setting

$$ \sigma_0:=
\|u\|_{L^\infty(\R^n)}
+\gamma_1+ \|(f+g)_-\|_{L^\infty(B_{4})}^{1/2}\,
\|u\|_{L^\infty(B_{4})}^{1/2},$$

we have that, for all $x\in B_2$ and $\rho\in(0,1)$,

\begin{equation*}\rho
\|\partial_e u\|_{L^\infty(B_{\rho}(x))}\le
\e \rho\,\|\partial_e u\|_{L^\infty(B_{3\rho}(x))}\end{equation*}

and accordingly

\begin{equation*}
\| u\|_{L^\infty(B_{\rho}(x))}+
\rho
\|\partial_e u\|_{L^\infty(B_{\rho}(x))}\le C_\e\,\sigma_0+
\e \rho\,\|\partial_e u\|_{L^\infty(B_{3\rho}(x))},\end{equation*}

up to renaming $C_\e$.
We can thereby apply
Lemma <ref> with $m:=1$ and deduce that

$$ \|\partial_e u\|_{L^\infty(B_{1/2})}\le C\sigma_0,$$

for a constant $C$ depending only on the quantities in the statement of Theorem <ref>.
The desired result follows from this estimate and a covering argument.

§.§ Proof of Theorem <ref>

The proof of Theorem <ref>
can be
carried out similarly to the one of Theorem <ref>. The main difference
is that here we only perform first derivative estimates and therefore the choice
of the test function must be modified accordingly.

The technical differences with respect to
the proof of Theorem <ref> can be listed as follows:

* $A_x\in{\mathcal{A}}$
must be replaced by $B_x\in{\mathcal{B}}$, as well as
the notation ${\mathcal{L}}_A$
must be changed into ${\mathcal{L}}_{K_B}$,

* The continuity of $g_A$ with respect to $A$ given by (<ref>)
is replaced here by the continuity of $g_B$ with respect to $B$, as given by (<ref>).
Also, here $u$ is a solution of the equation (<ref>) instead of the one in (<ref>).

* The analogue of the inequality (<ref>) involving the operator acting
on second order derivatives is not needed here,
since we focus on first derivative estimates,

* Also, we do not need to exploit the change of variables in
formula (<ref>) in this case,

* The use of Proposition <ref> is replaced here
by the use of Theorem <ref>, but then
the right-hand side of the analogue inequality to (<ref>)
presents an additional reminder
$\e^2\,\|\partial_e u\|^2_{L^\infty(B_3)}$,
and deals only with the first derivative
– more precisely,
inequality (<ref>) is replaced here
by (<ref>) of Theorem <ref>,

* The first derivative estimate in (<ref>)
presents an additional reminder $\e\,\|\partial_e u\|_{L^\infty(B_3)}$,
and from this one obtains the desired result by employing the
scaled estimates to reabsorb this remainder
in the left-hand side (via Lemma <ref>, used here with $m:=1$, and a covering argument),

* The second derivative estimate in (<ref>) is dropped in this case.

§.§ “Downstairs”
proof of the first key inequality for $s$-harmonic functions

The next result is a particular
case of Proposition <ref>.
While the proofs of Propositions <ref>
and <ref>
relied on extension methods, we provide here a
proof of the particular case
without using the extension. Note that, as in Propositions <ref>
and <ref>, $\sigma_0$ is independent of $s$.

Let $\eta\in C^\infty(\R^n)\cap W^{2,\infty}(\R^n)$,
and assume that $u\in W^{1,\infty}(\R^n)$
is a (weak) solution of

\begin{equation}\label{92i-9393}
(-\Delta)^s u =0 \quad{\mbox{ in }}B_1.
\end{equation}

Then, there exists $\sigma_0>0$,
depending only on $n$ and $\| \eta\|_{C^2(\R^n)}$, such that

\begin{equation}\label{10-10-XT6}
(-\Delta)^s \big( \eta^2|\nabla u|^2+\sigma u^2\big)\le 0\quad{\mbox{ in }}B_1\quad
\text{ if } \sigma\ge \sigma_0.
\end{equation}

The proof relies on several integrations by
parts, which carefully take into account oscillations
and compensations inside the integrals.

We observe that, by regularity results for (<ref>), $u\in C^\infty(B_1)$.
Exploiting Proposition <ref>, in order
to prove (<ref>),
it suffices to show, by translation invariance, that

\begin{equation}\label{01234-11-T6.3}\begin{split}&
\eta(0)-\eta(y)\big)\nabla u(0)\cdot\nabla u(y)\,K(y)\,dy\\ &\qquad\quad\leq\,
\int_{\R^n} \big|
\eta(0)\nabla u(0)-\eta(y)\nabla u(y)\big|^2\,K(y)\,dy+
\sigma\,\int_{\R^n}\big|u(0)-u(y)\big|^2\,K(y)\,dy
\end{split}
\end{equation}

knowing that

\begin{equation}\label{STARR}
(-\Delta)^s u (0)=0.
\end{equation}

To prove this, we call $I_1$ the left-hand side of (<ref>).
Integrating by parts, we have

\begin{equation}\label{01p10}
\begin{split}
\eta(0)-\eta(y)\big)\nabla u(0)\cdot\nabla \big(u(y)-u(0)\big)\,K(y)\,dy\\
-2\int_{\R^n}\eta(0)\,\nabla u(0) \big(u(y)-u(0)\big)\cdot
\nabla\Big( \big(
\eta(0)-\eta(y)\big)\,K(y)\Big)\,dy.
\end{split}
\end{equation}

We remark that, to obtain this integration by parts identity,
one must argue in balls $B_R$ and use that the boundary terms
on $\partial B_R$ go to zero as $R\to+\infty$ (as well as the integrability in $\R^n$ of the
above functions), thanks to
the decay of the kernel[As a matter of fact, though the statement
of Lemma <ref> is specific for the fractional Laplacian,
we perform the initial part of the proof arguing for a general kernel $K$,
to isolate the only point where we will use that $K(z)=c_{n,s}|z|^{-n-2s}$.]
and of its derivatives assumed in (<ref>) and (<ref>).

Moreover, from the bound (<ref>)
on the first derivative of the kernel,

\begin{equation}\label{RIAS} \Big|
\nabla\Big( \big(
\eta(0)-\eta(y)\big)\,K(y)\Big)\Big|\le
\big|\nabla\eta(y)\big|\,K(y)+
\big|
\eta(0)-\eta(y)\big|\,\big|\nabla K(y)\big|
\le CK(y),\end{equation}

where, from now on,
$C$ denotes different constants depending only on $n$ and $\|\eta\|_{C^2(\R^n)}$
(in particular, independent of $s$ in our case, that is, when
the kernel $K$ is that of the fractional Laplacian).
Therefore, we can write (<ref>) as

\begin{equation*}
\begin{split}
2\int_{\R^n}\big(\eta(y)\,\nabla u(y)-\eta(0)\,\nabla u(0)\big) \big(u(y)-u(0)\big)\cdot
\nabla\Big( \big(
\eta(0)-\eta(y)\big)\,K(y)\Big)\,dy\\&\quad
-2\int_{\R^n}\eta(y)\,\nabla u(y) \big(u(y)-u(0)\big)\cdot
\nabla\Big( \big(
\eta(0)-\eta(y)\big)\,K(y)\Big)\,dy
\\ &\le CI_2-2\int_{\R^n}\eta(y)\,\nabla u(y) \big(u(y)-u(0)\big)\cdot
\nabla\Big( \big(
\eta(0)-\eta(y)\big)\,K(y)\Big)\,dy,
\end{split}
\end{equation*}

\begin{equation*}
\int_{\R^n}\big|\eta(y)\,\nabla u(y)-\eta(0)\,\nabla u(0)\big| \big|u(y)-u(0)\big|
\end{equation*}

Now, integrating by parts and using again (<ref>), we have

\begin{equation*}
\begin{split}
& 2\left| \int_{\R^n}\eta(y)\,\nabla u(y) \big(u(y)-u(0)\big)\cdot
\nabla\Big( \big(
\eta(0)-\eta(y)\big)\,K(y)\Big)\,dy\right|\\&\qquad\quad
=\;\left| \int_{\R^n}\eta(y)\,\nabla \big|u(y)-u(0)\big|^2\cdot
\nabla\Big( \big(
\eta(0)-\eta(y)\big)\,K(y)\Big)\,dy\right| \\&\qquad\quad=\;\left|
\int_{\R^n}\big|u(y)-u(0)\big|^2\, {\rm div}\Big(\eta(y)
\nabla\Big( \big(
\eta(0)-\eta(y)\big)\,K(y)\Big)\Big)\,dy\right|\\&\qquad\quad \le\;
\int_{\R^n}\big|u(y)-u(0)\big|^2\,|\nabla\eta(y)|\,K(y)\,dy
\\&\qquad\quad\qquad\quad+\left|
\int_{\R^n}\big|u(y)-u(0)\big|^2\,\eta(y)\,
\Delta\Big( \big(
\eta(0)-\eta(y)\big)\,K(y)\Big)\,dy\right|\\ &\qquad\quad\le\;

\begin{equation*}
\end{equation*}

\begin{equation*}
\int_{\R^n}\big|u(y)-u(0)\big|^2\,\eta(y)\,
\nabla\eta(y)\cdot\nabla K(y) \,dy,\end{equation*}

\begin{equation}\label{DTT}
\int_{\R^n}\big|u(y)-u(0)\big|^2\,\eta(y)\,
\big(
\eta(0)-\eta(y)\big)\,\Delta K(y) \,dy.
\end{equation}

Clearly, $I_2$ and $I_3$ are “good terms” which are controlled by the
right-hand side of (<ref>). Hence, to bound $I_1$ it remains to control $|T_1|$ and $|T_2|$.

To estimate $T_1$, we observe that

\begin{equation*}
\big|\nabla\eta(y)-\nabla\eta(0)\big|\,|\nabla K(y)|
\le C\,|y|\,|\nabla K(y)|
\le CK(y),\end{equation*}

thanks to the bound (<ref>)
on the first derivative of the kernel, and therefore, integrating by parts,

\begin{equation*}
\begin{split}
\int_{\R^n}\big|u(y)-u(0)\big|^2\,\eta(y)\,
\nabla\eta(0)\cdot\nabla K(y) \,dy\right|\\&\qquad
\int_{\R^n}\big|u(y)-u(0)\big|^2\,\eta(y)\,
\big(\nabla\eta(y)-\nabla\eta(0)\big)\cdot\nabla K(y) \,dy\right|\\&\le\left|
\int_{\R^n}\nabla\Big(\big|u(y)-u(0)\big|^2\,\eta(y)\Big)\cdot
\nabla\eta(0)\, K(y) \,dy\right|
\int_{\R^n}\nabla\big|u(y)-u(0)\big|^2\,\eta(y)\cdot
\nabla\eta(0)\, K(y) \,dy\right|
+CI_3\\&\le 2\left|
\int_{\R^n}\big(u(y)-u(0)\big)\,\eta(y)\,\nabla u(y)\cdot
\nabla\eta(0)\, K(y) \,dy\right|
\int_{\R^n}\big(u(y)-u(0)\big)\,\eta(0)\,\nabla u(0)\cdot
\nabla\eta(0)\, K(y) \,dy\right|+C(I_2+I_3)\\
\end{split}
\end{equation*}

where (<ref>) has been used in the last line.

Finally, we estimate $T_2$. To this end, we take $\zeta\in C^\infty_c(B_1)$
with $\zeta=1$ in $B_{1/2}$, and we define

$$ T_3:=\int_{\R^n}\big|u(y)-u(0)\big|^2\,\eta(y)\,
\nabla\eta(0)\cdot y\,\zeta(y)\,\Delta K(y) \,dy.$$

by the second derivative bound (<ref>) on the kernel and (<ref>), we have

\begin{equation}\label{0203487ow023939934}
\begin{split}
\int_{\R^n}\big|u(y)-u(0)\big|^2\,\eta(y)\,
\big(
\eta(0)-\eta(y)+\nabla\eta(0)\cdot y\,\zeta(y)\big)\,\Delta K(y) \,dy-T_3\right|\\
&\le C
\int_{\R^n}\big|u(y)-u(0)\big|^2\,|y|^2\,|D^2 K(y)| \,dy
\end{split}\end{equation}

It only remains to bound $T_3$.
For this, when the kernel $K$ is that of the fractional Laplacian, we have that

\begin{eqnarray*}&& 2(s+1)\nabla K(y)+y\Delta K(y)\\&&\qquad=-
c_{n,s} \,(n+2s)\,y\,\left( \frac{2(s+1)}{|y|^{n+2s+2}}
+\sum_{i=1}^n \left( \frac1{|y|^{n+2s+2}}-\frac{(n+2s+2)\,y_i^2}{|y|^{n+2s+4}}\right)
\right)=0.\end{eqnarray*}

Consequently, using once more
the first derivative bound (<ref>)
on the kernel, and integrating again by parts, we find that

\begin{eqnarray*}
|T_3|&=& 2(s+1)\,\left|
\int_{\R^n}\big|u(y)-u(0)\big|^2\,\eta(y)\,
\nabla\eta(0)\,\zeta(y)\cdot\nabla K(y) \,dy\right|
\\&\le& 2(s+1)\,\left|
\int_{\R^n}\big|u(y)-u(0)\big|^2\,\eta(y)\,
\nabla\eta(0)\,\zeta(0)\cdot\nabla K(y) \,dy\right|\\&&\quad
\int_{\R^n}\big|u(y)-u(0)\big|^2\,\eta(y)\,|
\nabla\eta(0)|\,|y|\,|\nabla K(y)| \,dy
\\ &\le&
\int_{\R^n} \nabla\Big(\big|u(y)-u(0)\big|^2\,\eta(y)\Big)\cdot
\nabla\eta(0)\,\zeta(0)\, K(y) \,dy\right|
\\ &\le&
\int_{\R^n} \nabla \big|u(y)-u(0)\big|^2\,\eta(y)\cdot
\nabla\eta(0)\,\zeta(0)\, K(y) \,dy\right|
\int_{\R^n} \big(u(y)-u(0)\big)\,\eta(y)\,
\nabla u(y)\cdot
\nabla\eta(0)\,\zeta(0)\, K(y) \,dy\right|
\int_{\R^n} \big(u(y)-u(0)\big)\,\eta(0)\,
\nabla u(0)\cdot
\nabla\eta(0)\,\zeta(0)\, K(y) \,dy\right|
\\&&\quad+
\int_{\R^n} \big(u(y)-u(0)\big)\,\big(\eta(y)\,
\nabla u(y)-\eta(0)\,\nabla u(0)\big)\cdot
\nabla\eta(0)\,\zeta(0)\, K(y) \,dy\right|\\&&\quad
&\le & 0+C(I_2+I_3),
\end{eqnarray*}

where we have exploited (<ref>) once more in the last step.

§ A MAXIMUM PRINCIPLE IN $\R^{N+1}_+$

We give here an elementary proof of a maximum principle for the extension operator
in the whole halfspace.
The result may have appeared somewhere else, but we could not find a reference.

Let $V$ be bounded by above in $\R^{n+1}_+$ and
satisfy $L_a V\le0$ weakly[We could instead assume that $V$ is locally $C^{1,1}$
in $\R^{n+1}_+$ and that $L_aV\le0$ is satisfied at every point in $\R^{n+1}_+$ in the nondivergence
sense of Remark <ref> and Lemma <ref> (see also the comments after the lemma). The proof of
Lemma <ref> in this setting is the same as the one that we give below. Simply notice that the
maximum principle in the half ball ${\mathcal{B}}_{4R}^+$ also holds for this notion of
in $\R^{n+1}_+$. Assume that $V(x,0)\le0$
for $x\in\R^n$ (in the trace sense).

Then, $V\le0$ in $\R^{n+1}_+$.

Replacing $V$ by $V/\big(1+\|V^+\|_{L^\infty(\R^{n+1}_+)}\big)$, we can suppose that $
V\le1$ in $\R^{n+1}_+$.
For $R>0$, we define $
{\mbox{ s.t. }}|(x,y)|<R\}$ and $

Let $\overline{W}\in C^\infty(\R^{n+1})$ be nonnegative and
such that $\overline{W}=0$ in ${\mathcal{B}}_{2}$ and $\overline{W}=1$
in $\R^{n+1}\setminus {\mathcal{B}}_{3}$.
We define $W:\R^{n+1}_+\to\R$ to be the minimizer of the extended
Dirichlet energy
\int_{{\mathcal{B}}_4^+} y^a|\nabla U(x,y)|^2\,dx\,dy$$

among the functions $U$ having finite energy and such
that $U=\overline{W}$ along $\partial{\mathcal{B}}_4^+$.
We point out that this is a well posed problem in the sense
of Sobolev spaces with Muckenhoupt  weights
(which indeed possess a suitable notion of trace);
see <cit.> and Theorem 3.2 of <cit.>.

Now, for every $R>0$, we let $W_R(x,y):=W\left(\frac{x}R,\frac{y}R\right)$. We note that $W_R(x,y)=\overline{W}\left(\frac{x}R,\frac{y}R\right)=1\ge V(x,y)$
for $(x,y)\in\partial {\mathcal{B}}_{4R}\cap \R^{n+1}_+$.
Furthermore, $W_R(x,0)=W\left(\frac{x}R,0\right)
=\overline{W}\left(\frac{x}R,0\right)\ge0\ge V(x,0)$ for $x\in B_{4R}$.
As a consequence, by the maximum
principle (see <cit.>) we infer

\begin{equation}\label{7Co7}
V(x,y)\le W_R(x,y)=W\left(\frac{x}R,\frac{y}R\right)\quad{\mbox{
for }} (x,y)\in{\mathcal{B}}_{4R}^+.\end{equation}

We also define $W_o:{\mathcal{B}}_2\to\R$
to be the odd reflection of $W$ in the variable $y$. Thus,
we have that $-{\rm div}(|y|^a\nabla W_o)=0$ weakly
in ${\mathcal{B}}_2$.
Hence, we can apply the results in <cit.>
(or Theorem 3.3 of <cit.>)
and deduce that $W_o$ is
Hölder continuous, and thus so is $W$ up to the boundary $B_{2R}\times\{0\}$.
From this and (<ref>)
the desired result
follows by sending $R\to+\infty$.

§ SCALED INEQUALITIES

We recall a classical result about scaled inequalities which we have
used to prove Theorems <ref>
and <ref>.

Let $m\ge0$ be an integer and $u\in C^m(B_5)$, with $B_5\subset\R^n$.
Let $\sigma_0\ge0$, and assume that for every $\e>0$ there exists a constant $C_\e\ge0$
such that

\begin{equation}\label{0380392410}
\sum_{k=0}^m \rho^k \| D^k u\|_{L^\infty(B_\rho(x))}
\le C_\e \,\sigma_0+\e\sum_{k=0}^m
\rho^k\| D^k u\|_{L^\infty(B_{3\rho}(x))}
\end{equation}

for every $x\in B_2$ and every $\rho\in(0,1)$.

Then, there exists a constant $C$, depending[It is important to
notice that, in the statement of Lemma <ref>,
the constant $C$ does not depend only
on $m$ —but on $m$ and on the value of the
function $C_\varepsilon=:C(\varepsilon)$
at $\epsilon:=\big(2(m+1)4^{m+1}\big)^{-1}$.
When we apply Lemma <ref>
to prove Theorem <ref>, this observation is crucial
in order to obtain the correct dependencies of the structural constants
in Theorem <ref> (instead of obtaining a universal constant).
]only on $m$
and on the value of the constant $C_\epsilon$
with $\epsilon:=\big(2(m+1)4^{m+1}
\big)^{-1}$, such that

\begin{equation}\label{9eudf3331144}
\sum_{k=0}^m \| D^k u\|_{L^\infty(B_{1/2})}\le C \sigma_0.
\end{equation}

\begin{eqnarray*}&&
S(x):=\sum_{k=0}^m (1-|x|)^{k+1}\,|D^ku(x)|\qquad\quad
{\mbox{ and }}\qquad \quad M:=\max_{x\in\overline{B}_1} S(x).
\end{eqnarray*}

Since $S$ vanishes along $\partial B_1$, the maximum of $S$
in $\overline{B}_1$ is attained at an interior point: namely, there exists $x_0$
with $|x_0|<1$ such that $S(x_0)=M$.

We set

$$ \rho_0:=\frac{1-|x_0|}{4}\in(0,1).$$

Then, we have that $B_{3\rho_0}(x_0)\subset B_1$,
and, for all $x\in B_{3\rho_0}(x_0)$, it holds that $1-|x|\ge1-|x_0|-3\rho_0=\rho_0$.
As a consequence,

\begin{eqnarray*}&& \sum_{k=0}^m \rho_0^k\| D^k u\|_{L^\infty(B_{3\rho_0}(x_0))}\leq
\sum_{k=0}^m \rho_0^k \sup_{x\in B_{3\rho_0}(x_0)}
\frac{S(x)}{(1-|x|)^{k+1}}\\&&\qquad\qquad
\le \rho_0^{-1}\sum_{k=0}^m \sup_{x\in B_{3\rho_0}(x_0)} S(x)
\le  \rho_0^{-1} (m+1)M.\end{eqnarray*}

From this and (<ref>), we infer that

\begin{eqnarray*}
C_\e \,\sigma_0+\e (m+1)M\ge\rho_0
C_\e \,\sigma_0+\rho_0\e\sum_{k=0}^m
\rho_0^k\| D^k u\|_{L^\infty(B_{3\rho_0}(x_0))}
\ge\rho_0
\sum_{k=0}^m \rho_0^k \| D^k u\|_{L^\infty(B_{\rho_0}(x_0))}
\\ &&\quad\ge
\sum_{k=0}^m
\frac{(1-|x_0|)^{k+1}}{4^{k+1}}\, | D^k u(x_0)|
\ge
\sum_{k=0}^m
\frac{(1-|x_0|)^{k+1}}{4^{m+1}}\, | D^k u(x_0)|
=\frac{S(x_0)}{4^{m+1}}=\frac{M}{4^{m+1}} .\end{eqnarray*}

Therefore, taking $\e:=\frac{1}{2(m+1)\,4^{m+1}}$, we
conclude that $S\le M\le C\sigma_0$ in $B_1$ for some constant $C$ depending only on $m$
and on the constant $C_\epsilon$
with $\epsilon:=\big(2(m+1)4^{m+1}
\big)^{-1}$.
From this, the bound in (<ref>), in half the ball, follows immediately.

§ A VARIANT OF PROPOSITION <REF>

We state and prove here a convenient modification of Proposition <ref>
that is used in the proof of Lemma <ref>, in which $\partial_eu$
is replaced by $\nabla u$.

Let $K$ satisfy (<ref>)
and (<ref>), and let ${\mathcal{L}}_K$ be defined by (<ref>). Given a function $u\in C^\infty(\R^n)\cap W^{1,\infty}(\R^n)$,
$\eta\in C^\infty(\R^n)\cap L^{\infty}(\R^n)$, and $\CALERRE\in\R$, consider

\begin{equation*} \varphi:=\eta^2 |\nabla u|^2 +\sigma u^2.\end{equation*}

Then, the inequality

\begin{equation*}
{\mathcal{L}}_K \varphi\le 2\eta^2\,\nabla u\cdot{\mathcal{L}}_K \nabla u+2\sigma u\,{\mathcal{L}}_K u+\CALERRE
\end{equation*}

holds at a point $x\in\R^n$ if and only if

\begin{equation*}
\begin{split}&
2\int_{\R^n}\eta(x)\,\big( \eta(x)-\eta(y)\big)\,\nabla u(x)\cdot\nabla u(y)\,K(x-y)\,dy
\\ &\qquad\le\,\int_{\R^n}\big|\eta(x)\,\nabla u(x)-\eta(y)\,\nabla u(y)\big|^2\,K(x-y)\,dy
\\&\qquad\qquad+\sigma\int_{\R^n}\big|u(x)- u(y)\big|^2\,K(x-y)\,dy

We observe that

\begin{eqnarray*}
&& {\mathcal{L}}_K \Big( \eta^2\,|\nabla u|^2+\sigma
\,u^2\Big)(x)
-2\eta^2(x)\, \nabla u(x)\cdot{\mathcal{L}}_K\,\nabla u (x)
\\&&\qquad\qquad-2\sigma\,u(x)\,{\mathcal{L}}_K u(x)+\sigma\int_{\R^n}\big|u(x)-u(y)\big|^2\,K(x-y)\,dy\\
&&\qquad= \int_{\R^n}
\Big( \eta^2(x)\,|\nabla u(x)|^2
-\eta^2(y)\,|\nabla u(y)|^2\Big)\,K(x-y)\,dy\\&&\qquad\qquad
+\sigma\int_{\R^n} \big(u^2(x)-u^2(y)\big)\,K(x-y)\,dy\\
&&\qquad\qquad-2\eta^2(x)\, \nabla u(x)\cdot
\int_{\R^n}\Big( \nabla u(x)-\nabla u(y)\Big)
\,K(x-y)\,dy\\
&&\qquad\qquad-2\sigma\,u(x)\,\int_{\R^n} \big(u(x)-u(y)\big)\,K(x-y)\,dy
\\
&&\qquad= \int_{\R^n}\Big( 2\eta^2(x)\, \nabla u(x)\cdot\nabla u(y)
-\eta^2(x)\, |\nabla u(x)|^2-
\eta^2(y)\, |\nabla u(y)|^2
\Big)\,K(x-y)\,dy
\\ &&\qquad=2\int_{\R^n} \eta(x)\,\big(\eta(x)-\eta(y)\big)\, \nabla u(x)
\cdot\nabla u(y)
\,K(x-y)\,dy
\\&&\qquad\qquad-
\int_{\R^n}\Big|
\eta(x)\, \nabla u(x)-
\eta(y)\, \nabla u(y)
\Big|^2\,K(x-y)\,dy.
\end{eqnarray*}

From this, the proposition follows readily.

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