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.

What is $(-1)^{\sqrt{2}}$? Is $x \mapsto x^{\sqrt{2}}$ continuous on the reals? — PhoemueX, Dec 28 '22 at 07:18
You need to write $|u|^r$. For continuity use dominated convergence. — daw, Dec 28 '22 at 15:06

score 2 · Accepted Answer · answered Jan 01 '23 at 14:57

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.

Thank you Davide Giraudo for the reply. – Lucas Linhares Jan 05 '23 at 14:21 — Lucas Linhares, Jan 05 '23 at 14:21
@Thiago You are welcome. – Davide Giraudo Jan 05 '23 at 15:40 — Davide Giraudo, Jan 05 '23 at 15:40

On the continuity of Nemytskii operators

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