On the continuity of Nemytskii operators

Question

Let $\Omega \subset \mathbb{R}^N$ and for $r,p > 1$ consider $$T: L^{pr}(\Omega) \to L^p(\Omega), \qquad T(u) = |u|^r.$$ I need to prove that $T$ is continous, but I don't know how to proceed. I have seem that this comes from the Nemytskii operators theory, but I couldn't find a reference with this proof.

Answer 1

Let $u\in\mathbb L^{pr}(\Omega)$. We have to prove that if $(u_n)$ is a sequence converging in $L^{pr}(\Omega)$, then $\lVert T(u_n)-T(u)\rVert_{\mathbb L^p(\Omega)}\to 0$. Note that it suffices to do it for $(u_n)$ satisfying $u_n\to u$ in $\mathbb L^{pr}(\Omega)$ and almost everywhere.

We use the following result: if $(v_n)$ is a sequence in $\mathbb L^q(\Omega)$ such that $v_n\to v$ almost everywhere and $\lVert v_n\rVert_{\mathbb L^q(\Omega)}\to \lVert v\rVert_{\mathbb L^q(\Omega)}$, then $\lVert v_n-v\rVert_{\lVert u_n\rVert_{\mathbb L^q(\Omega)}}\to 0$.

Applying this with $q=pr$, $v_n=\lvert u_n\rvert^r$ and $v=\lvert u\rvert^r$ gives the result.