Fixed point $u(x) = F(u)(x)$ properties

Question

Let $F:H^1(\Omega) \to H^1(\Omega)$ be some map. Suppose we have a function $u$ such that for some $x \in \Omega$, $$u(x) = F(u)(x).$$

From this, I think it follows that $$u(x) = F(F(...F(u))..)(x) = F^{(n)}(u)(x)$$ (that's composition $n$ times) for any $n$.

This seems a bit strange to me. Can we say anything about $F$ or $u$ if this happens for $x \in D \subset \Omega$ with $D$ not a null set?

Answer 1

This conclusion is not valid. Take $\Omega =\mathbb R$, $F$ defined by $(Fu)(x) = u(x+1)$.

Take $u$ which is zero outside of $(-1,2)$ and one in $[0,1]$. Set $x=0$. Then $$ u(0)=1, \ F(u)(0) = 1, \ F(F(u))(0)=0. $$