If $u(x) = T(u)(x)$ on a subset and $T$ is Lipschitz is $u(x) = T(T(u))(x)$?

Question

Let $T:H^1(\Omega) \to H^1(\Omega)$ be a Lipschitz continuous map. Let $\Omega_0 \subset \Omega$ be a subdomain contained in $\Omega$.

Suppose we are given a function $u \in H^1(\Omega)$ such that for all $x \in \Omega_0$, $$u(x) = T(u)(x).$$ Could it be that $u(x) = T(T(u))(x)$ whenever $x \in \Omega_0$?

It's not true I think when the first identity holds for an arbitrary point, but if it holds for a non-null set and we have continuity it may hold??

Answer 1

Let $T (v)(x) = \int_\Omega v(t) dt - 1$ (a constant function). Let $u$ be a function which is zero in $\Omega_0$ with $\int_\Omega u =1$. Then

$$Tu(x) = u(x) = 0$$

on $\Omega_0$. But

$$T(T(u))(x) = -1$$

does not equal $u$ on $\Omega_0$.