If $u_n \to u$ in $L^2(0,T;L^2(\Omega))$ and $f_n \to f$ uniformly, does $f_n(u_n) \to f(u)$ in $L^2(0,T;L^2(\Omega))$?

Question

Let $\Omega$ be an unbounded domain. Suppose we have $u_n \to u$ in $L^2(0,T;L^2(\Omega))$.

Let $f_n\colon \mathbb{R} \to \mathbb{R}$ be a sequence such that $f_n \to f$ uniformly. We know that $f_n(u_n)$ and $f(u)$ are in $L^2(0,T;L^2(\Omega))$. And $f_n$ and $f$ are both continuous functions.

Does it follow that $f_n(u_n) \to f(u)$ in $L^2(0,T;L^2(\Omega))$?

Unfortunately, the Lipschitz constant of $f_n$ depends on $n$ in a bad way, so i cannot use the DCT. Does anyone know a better idea? Maybe I can use Nemytskii operator in some way.

Answer 1

This is not true. Consider $\Omega = (0,1)$ and $$u_n(x) = n^{1/4} \, \chi_{(0,1/n)}(x).$$ Then $u_n \to 0$ in $L^2(\Omega)$. But for $f(x) = x^2$, you have $$f(u_n(x)) = n^{1/2} \, \chi_{(0,1/n)}(x),$$ which has norm $1$. Hence $f(u_n)$ does not converge towards $f(u)$.

One does not even need a varying $f_n$.