since $1\leq2(|\xi_1|+\dots+|\xi_r|+\ell_1+\dots+\ell_s)$, so the claim follows.

Let $L$ be a differential operator on $\G$ with constant coefficients, with symbol $\sigma_L$, that is, $\sigma_L:\mathbb{Z}^{r+1}\times\frac{1}{2}\Z^s\to\mathbb{C}$ such that  for any $f\in \mathscr{D}'(\G)$:

$$\widehat{Lf}(\tau,\xi,\ell)_{\alpha\beta} = \sigma_L(\tau,\xi,\alpha)\widehat{f}(\tau,\xi,\ell)_{\alpha\beta}$$

$$\widehat{\Lt f}(\tau,\xi,\ell)_{\alpha\beta}=\sigma_L(-\tau,-\xi,-\alpha)\widehat{f}(\tau,\xi,\ell)_{\alpha\beta}$$

for each $(\tau,\xi)\in\Z^{r+1}, \ell\in\frac{1}{2}\N_0^s, -\ell\leq\alpha,\beta\leq \ell$. Then

$$(\ker \Lt)^0= \left\{f\in \mathscr{D}'(\G)| \text{ such that } \widehat{f}(\tau,\xi,\ell)_{\alpha\beta} = 0 \text{ if } \sigma_L(\tau,\xi,\alpha)=0\right\}.$$

Suppose first that $\widehat{f}(\tau,\xi,\ell)_{\alpha\beta} = 0 \text{ if }\sigma_L(\tau,\xi,\alpha)=0$. Let $v\in C^\infty(\G)$ be such that $v\in\ker \Lt$. Then, by Lemma <ref>:

\begin{align*}
\langle f,v\rangle &= (2\pi)^{r+1}\sum_{(\tau,\xi)\in\Z^{r+1}}\sum_{\ell\in\frac{1}{2}\N_0^s}d_l\sum_{-\ell\leq\alpha,\beta\leq \ell}\widehat{f}(\tau,\xi,\ell)_{\alpha\beta}\widehat{v}(-\tau,-{\xi},\ell)_{(-\alpha)(-\beta)}(-1)^{\sum \beta_j-\alpha_j}.
\end{align*}

If $\sigma_L(\tau,\xi,\alpha) = 0$, then $\widehat{f}(\tau,\xi,\ell)_{\alpha\beta} = 0$. If not, then since

\begin{align*}
0 &= \widehat{\Lt v}(-\tau,-{\xi},\ell)_{(-\alpha)(-\beta)} = \sigma_L(\tau,\xi,\alpha)\widehat{v}(-\tau,-{\xi},\ell)_{(-\alpha)(-\beta)}
\end{align*}

this implies $\widehat{v}(-\tau,-{\xi},\ell)_{(-\alpha)(-\beta)}=0$, so every term in the sum above is zero and $\langle f,v\rangle=0$, so that $f\in(\ker \Lt)^0$. Now let $f\in(\ker \Lt)^0$ and suppose $\sigma_L(\tau,\xi,\alpha)=0$. For each $-\ell\leq\beta\leq \ell$, $l-\beta\in\N_0^s$, take $v_{\tau,\xi,\ell,\alpha,\beta}\in C^{\infty}(\G)$ given by:

$$ \widehat{v_{\tau,\xi,\ell,\alpha,\beta}}(-\tau,-\xi,\ell)_{(-\alpha)(-\beta)} = 1,$$

And $\widehat{v_{\tau,\xi,\ell,\alpha,\beta}}(\tau',\xi
',\ell')_{\alpha'\beta'}=0$ otherwise.

\begin{align*}
\widehat{\Lt v_{\tau,\xi,\ell,\alpha,\beta}}(-\tau,-{\xi},\ell)_{(-\alpha)(-\beta)} &= \sigma_L(\tau,\xi,\alpha)\widehat{v_{\tau,\xi,\ell,\alpha,\beta}}(-\tau,-{\xi},\ell)_{(-\alpha)(-\beta)}\\
\end{align*}

so $v_{\tau,\xi,\ell,\alpha,\beta}\in\ker \Lt$. Therefore, by Lemma <ref>:

$$0 = \langle f,v_{\tau,\xi,\ell,\alpha,\beta}\rangle = d_l\widehat{f}(\tau,\xi,\ell)_{\alpha\beta}(-1)^{\sum\beta_j-\alpha_j}$$

so we conclude the other inclusion also holds.

Let $g,\theta \in C^{\infty}(\T^1)$ and $\theta_0 \doteq \frac{1}{2\pi}\int_0^{2\pi}\theta(t)\mathop{dt}$. If $\theta_0\not\in i\mathbb{Z}$, then the differential equation

\begin{equation}\label{ode}
\partial_t u(t)+\theta(t)u(t)=g(t),\,\qquad t\in\T^1
\end{equation}

admits unique solution in $C^\infty(\T^1)$ given by:

\begin{equation}\label{sol-}
u(t) = \frac{1}{1-e^{-2\pi\theta_0}}\int_0^{2\pi}g(t-s)e^{-\int_{t-s}^t\theta(\tau)d\tau}ds
\end{equation}

or equivalently by:

\begin{equation}\label{sol+}
u(t) = \frac{1}{e^{2\pi\theta_0}-1}\int_0^{2\pi}g(t+s)e^{\int_{t}^{t+s}\theta(\tau)d\tau}ds.
\end{equation}

If $\theta_0\in i\mathbb{Z}$, then equation <ref> admits infinitely many solutions given by:

\begin{equation}\label{solz}
u_\lambda(t) = \lambda e^{-\int_0^t\theta(\tau)d\tau}+\int_0^t g(s)e^{-\int_s^t\theta(\tau)d\tau}ds
\end{equation}

for every $\lambda\in\mathbb{R}$, if and only if

$$ \int_0^{2\pi}g(t)e^{\int_0^t\theta(\tau)d\tau}\mathop{dt}=0.$$

Since the functions on the torus may be seen as $2\pi$ periodic functions on $\R$, the proof follows from simple differentiation of the formulas above and applying the periodicity.

Let $\phi\in C^\infty(\T^1)$ be a non-null function, and let $\Phi$ be a function such that $\Phi'=\phi$. Suppose there exists $m\in\R$ such that the sublevel set

$$\Omega_m = \{t\in\T^1; \Phi(t)<m\}$$

is not connected.

Then, there exists $m_0<m$ such that $\Omega_{m_0}$ has two connected components with disjoint closures. Consequently, we can define functions $g_0,v_0\in C^{\infty}(\T^1)$ such that:

$$\int_0^{2\pi}g_0(t)\mathop{dt}=0,\ \supp(g_0)\cap\Omega_{m_0}=\varnothing, \  \supp(v_0')\subset\Omega_{m_0}\text{ and } \int_0^{2\pi} g_0(t)v_0(t)\mathop{dt}>0.$$

Let $C_1\subset \T^1$ be a connected component of $\Omega_m$. Notice that $C_1$ is homeomorphic to an open interval and has two distinct boundary points: $\partial C_1=\{t_1,t_2\}$. Choose $t_3\in C_1$ such that $\Phi(t)<m$. Since $\Omega_m$ is not connected, there exists another connected component $C_2$ of $\Omega_m$ such that $C_1\cap C_2=\emptyset$. Similar to $C_1$, the component $C_2$ is also homeomorphic to an open interval and its boundary is given by two distinct points: $\partial C_2=\{t_4,t_5\}$. Choose $t_6\in C_2$ such that $\Phi(t_6)<m$.

Now, choose $\epsilon>0$ such that $m_0 \doteq \max\{\Phi(t_3),\Phi(t_6)\}+\epsilon<r$. Since $\Phi(t_1)=m$, by the continuity of $\Phi$, there exists an open set $U_1\subset \T^1$ containing $t_1$ such that $\Phi(t)>m_0$ for each $t\in U_1$. Similarly, we can find an open set $U_2$ containing $t_2$ with the same property.

Let $I$ and $J$ be the connected components of $\Omega_{m_0}$ that contain $t_3$ and $t_6$, respectively. It is important to note that $U_1$ and $U_2$ are contained in $\T^1\backslash(I\cup J)$. Moreover, $I\subset C_1$ and $J\subset C_2$ are “separated" by $U_1$ and $U_2$, which implies that their closures do not intersect. In other words, if $x\in \overline{I}\cap\overline{J}$, then there exist sequences $(x_n)_n\subset I$ and $(y_n)_n\subset J$ such that $x_n\to x$ and $y_n\to x$. However, since $x_n\in I\subset C_1$, it follows that $\Phi(x_n)< m_0$ for all $n$, which implies $\Phi(x)\leq m_0<m$. Therefore, we have $x\in C_1$. The same logic applies to $y_n$, $J$, and $C_2$, which leads to $x\in C_1\cap C_2$, which is a contradiction.

Let us consider the previously defined set as contained in the interval $K = [t_1,t_1+2\pi]\subset \R$. Without loss of generality, we can assume that

$$t_1<t_3<t_2\leq t_4<t_6<t_5\leq t_1+2\pi$$

$$t_3\in I\subset C_1=(t_1,t_2), \quad t_6\in J\subset C_2=(t_4,t_5)$$

$$U_1 = [t_1,t_1+\epsilon')\cup (t_1+2\pi-\epsilon',t_1+2\pi]$$

where $0<\epsilon'$ and $t_1+\epsilon'<t_3$ and

$$U_2 = (t_2-\epsilon'',t_2+\epsilon'')$$

where $0<\epsilon''$ and $t_3<t_2-\epsilon''<t_2+\epsilon''<t_6$.

Now, for $j=1,2$, let $g_j\in C_c^\infty(U_j)$ be a bump function such that $\int_0^{2\pi}g_j(t)\mathop{dt}=1$. Set $g_0 = g_2-g_1$, so that $\supp(g_0)\subset U_1\cup U_2$ and so $\supp(g_0)\cap \Omega_{m_0}=\emptyset$. Also,

Finally, let $\delta>0$ be such that $t_3+\delta\in I$ and $t_6-\delta\in J$. Choose $v_0\in C_c^{\infty}((t_3,t_6))$ such that $v_0\equiv1$ in $[t_3+\delta,t_6-\delta]$. In this case,

and $\supp(v_0')\subset I\cup J\subset\Omega_{m_0}$.

Let $\psi\in C^{\infty}(\T^1)$ be a smooth real function such that $\psi(s)\geq 0$ for all $s$, and let $s_0\in\T^1$ be	 a zero of order greater than one for $\psi$, i.e., $\psi(s_0)=0=\psi'(s_0)$. Then, there exists $M>0$ such that for all $\lambda>0$ sufficiently large and $\delta>0$:

$$\int_{s_0-\delta}^{s_0+\delta}e^{-\lambda\psi(s)}ds\geq \left(\int_{-\delta}^{\delta}e^{-s^2}ds\right)\lambda^{-1/2}M^{-1/2}.$$

Let us consider the Taylor expansion of $\psi$ around $s_0$. For each $s\in (s_0-\delta,s_0+\delta)$, there exists $s'\in (s_0-\delta,s_0+\delta)$ such that

$$\psi(s) = \frac{\psi''(s')}{2}(s-s_0)^2$$

Let $\tilde{M}=\sup_{s\in [s_0-\delta,s_0+\delta]}\left|\frac{\psi''(s)}{2}\right|\geq0$. If $\tilde{M}=0$, then $\psi\equiv0$ and the inequality is trivial with $M=1$. Otherwise, let $M=\tilde{M}$ and then for $\lambda M>1$ we have:

\begin{align*}
\int_{s_0-\delta}^{s_0+\delta}e^{-\lambda\psi(s)}ds&\geq \int_{s_0-\delta}^{s_0+\delta}e^{-(\sqrt{\lambda M}(s-s_0))^2}ds\geq \frac{1}{\sqrt{\lambda M}}\left(\int_{-\delta\sqrt{\lambda M}}^{\delta\sqrt{\lambda M}}e^{-s^2}ds\right) \\
&\geq \frac{1}{\sqrt{\lambda M}}\left(\int_{-\delta}^{\delta}e^{-s^2}ds\right).
\end{align*}

Let $\phi\in C^\infty(\T^1)$ be such that $\int_0^{2\pi}\phi(t)\mathop{dt} = 0$ and for every $r\in\R$, the set: $\Omega_r = \left\{t\in\T^1|\int_0^t\phi(\tau)d\tau<r\right\}$ is connected. Then so is the set:

$$\tilde{\Omega}_r = \left\{t\in\T^1|\int_0^t\phi(\tau)d\tau\geq r\right\}=\left\{t\in\T^1|-\int_0^t\phi(\tau)d\tau\leq -r\right\}.$$

This follows from $\tilde{\Omega}_r=\T^1\backslash\Omega_r$ and the general fact that any $A\subset\T^1$ is connected if and only if $\T^1\backslash A$ is connected. To see this, let $A$ be connected. Note that the claim $\T^1\backslash A$ is connected is trivially true if $A=\T^1$. Otherwise, then $\T^1\backslash A$ contains at least one point, which, without loss of generality we may assume it is $0=0+2\pi\Z$. If we consider

\begin{align*}
x&\mapsto x+2\pi\Z
\end{align*}

then $f$ is an homeomorphism. Since $A$ is connected, $I=f^{-1}(A)$ also is. But then $I$ is an interval, so $I^c = (0,2\pi)\backslash I$ is either an interval containing $(0,\epsilon)$ or $(2\pi-\epsilon,2\pi)$ for some $\epsilon>0$ or the disjoint union of two intervals containing $(0,\epsilon)\cup(2\pi-\epsilon,0)$, for some $\epsilon>0$. In both cases, since $\T^1\backslash A = f(I^c)\cup\{0+2\pi\Z\}$ it is clearly connected. Switching the roles of $A$ and $A^c$, the converse also follows.

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