Is there a function $u$ with $g(u)$ regular whose "truncation" $g(T(u))$ is not regular?

Question

I hope the title is not to misleading.

Assume you have a continuous function $g:R\to R$ and a Lebesgue measurable function $u:\Omega\to R$ for some bounded domain $\Omega$ such that $g(u)\in H^1(\Omega)$. Let now $T_k$ be the truncation function at level $k$, i.e. $$ T_k(u) =\begin{cases}k & u>k \\ u& u\leq k \end{cases} $$

The question that comes up is the following one. Does from $g(u) \in H^1(\Omega)$ also follow $g(T_k(u)) \in H^1(\Omega)$.

This is easy for $g$ being an increasing function, since then $g(T_k(u)) = T_{g(k)}g(u)$ and $T_k$ is piecewiese smooth.

Maybe this is a standard example. In that case I don't know it :) I haven't yet succeeded in proving a negative result on the question. Any ideas?

Answer 1

This is not true. Take $\Omega = (0,1)$, $u = \chi_{(0,1/2)}$ and $g(t) = t \, (1-t)$.

Then, $g(u) \equiv 0$, but $g(T_{1/2}(u))$ is discontinuous, hence not in $H^1(\Omega)$.