Let $(\Omega, \mathcal F, \mu)$ be a $\sigma$-finite measure space. Let $p, q \in [1, \infty)$. Let $a:\mathbb R \to \mathbb R$ be continuous such that $$ |a(x)| \le C |x|^{p/q} \quad \forall x \in \mathbb R, \quad (\star) $$ for some finite $C>0$. We consider a non-linear map $$ A:L^p(\Omega) \to L^q(\Omega), u \mapsto a \circ u. $$
Clearly, $A$ is well-defined by $(\star)$. I'm trying to prove
Theorem $A$ is continuous.
- Could you have a check on my below attempt?
- Is there a more direct approach?
Proof Let $u, u_n \in L^p(\Omega)$ such that $u_n \to u$ in $L^p$. We want to prove that $A (u_n) \to A (u)$ in $L^q$. Let $\varphi$ be a sub-sequence of $\mathbb N$. Let $f_n := A(u_{\varphi (n)})$ and $f:=A (u)$. It suffices to prove that $(f_n)$ has a sub-sequence that converges to $f$ in $L^q$. There is a sub-sequence of $\psi$ of $\mathbb N$ such that $u_{\psi (n)} \to u$ a.e. We will prove that $f_{\psi (n)} \to f$ in $L^q$. Because $a$ is continuous, $f_{\psi (n)} \to f$ a.e. Let $g_n := C^{1/p} u_n$. Then
- $g_{\varphi \circ \psi (n)} \xrightarrow{n \to \infty} u$ a.e.,
- $g_{\varphi \circ \psi (n)} \xrightarrow{n \to \infty} u$ in $L^p$, and
- $|f_{\psi (n)}|^q \le |g_{\varphi \circ \psi (n)}|^p$ for all $n$.
The claim then follows from below result, i.e.,
Lemma Let $(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space and $(E, |\cdot|)$ a Banach space. Let $p, q \in [1, \infty)$. Let $f,f_n \in \mathcal L_p (X, \mu, E)$ and $g,g_n \in \mathcal L_q (X, \mu, E)$ such that $f_n \to f$ a.e. and $g_n \to g$ a.e. Assume $|f_n|^p \le |g_n|^q$ a.e. for all $n$ and $\|g_n\|_q \xrightarrow{n \to \infty} \|g\|_q$. Then $\|f_n -f\|_p \xrightarrow{n \to \infty} 0$.
