Composition with a continuous map preserves $L^p$ convergence?

Question

Assume we have $f_n \rightarrow f$ in $L^p(U)$, where $U$ is a bounded set of an Euclidean space. Is it true that for any continuous $F$, $F(f_n) \rightarrow F(f)$ in $L^p(U)$ as well? It is clear when $F$ is Lipschitz.

Probably this is simple, but I could not find a counterexample yet.

Answer 1

Let $$f_n = n\chi_{\left[0,\frac{1}{n^{p+1}}\right]}$$ and $$F(x) = x^{\frac{p+1}{p}}.$$ Then $\|f_n\|_{L^p} = n^{-\frac{1}{p}}\rightarrow 0$, but $\|F(f_n)\|_{L^p} = 1$ for all $n$.