Let $A=\begin{bmatrix}2&1\\ 1&1\end{bmatrix}$ and notice it acts on the torus $T=\mathbb{R}^2/\mathbb{Z}^2$ via left-multiplication. It also acts on $C^0(T)$ via $(Af)(x):= f(Ax)$. We can extend this action to distributions $\mathcal{D}'(T) = (C^\infty(T))^*$ via $(Au)(g) = u(A^{-1}g)$ for any $g\in C^\infty(T)$; this agrees with the definition for continuous functions because
$$ \int_T f(x)g(A^{-1}x) dx = \int_T f(Ax)g(x) dx $$
by using a linear change of variables and using that $\det(A)=1$. How can we prove that if $f\in L^1(T)$ satisfies $Af=f$ then $f$ must be almost-everywhere constant, but that there exists an infinite-dimensional space of distributions $u$ satisfying $Au=u$?
For the latter, I have no idea how to begin. But for the former, I have the following approach: if $f$ were non-constant then for some disjoint open balls $B_1,B_2\subset\mathbb{C}$ we would have that $S_1:=\{ x : f(x)\in B_1\}\subset T$ and $S_2:=\{ x : f(x)\in B_2\}\subset T$ both have positive measure. Then since $f(Ax)=f(x)$ for all $x$, we obtain that $(A^n(S_1))\cap S_2=\varnothing$ for al $n\ge 0$. So we want to prove that the action of $A$ on $T$ is "ergodic" enough to mix everything. Here is where I get stuck.
Thank you for your help!