How to prove the following statement:

Given $f\in C(\mathbb{R})$ which is bounded and $\Omega\subset\mathbb{R}^n$ is a bounded domain, then for any $u\in L^2(\Omega)$, we have $f(u)\in L^2(\Omega)$ and $f$ is a continuous mapping from $L^2(\Omega)$ to $L^2(\Omega)$ in the sense that

$u\to f\circ u$ is continuous on $L^2(\Omega)$

  • $\begingroup$ Hint: on a bounded domain, bounded functions are $L^2$ $\endgroup$ – Glougloubarbaki Feb 6 '18 at 9:02
  • $\begingroup$ "$f$ is a continuous mapping from $L^2(\Omega)$ to $L^2(\Omega)$." That doesn't make sense. I think you mean to say "$u\to f\circ u$ is continuous on $L^2(\Omega)$" $\endgroup$ – zhw. Feb 6 '18 at 9:06
  • $\begingroup$ Yes you're right. I will modify the question later $\endgroup$ – whereamI Feb 6 '18 at 9:22

Due to the boundedness assumptions, $f(u)$ is in $L^p(\Omega)$ for all $p\in[1,+\infty]$.

Let $(u_n)$ converge to $u$ in $L^2(\Omega)$. Choose pointwise converging subsequence $(u_{n_k})$. Then $f(u_{n_k})$ converges pointwise to $f(u)$. In addition we have the integrable upper bound $$ |f(u)(x) - f(u_{n_k})(x)|^2 \le 4M^2, $$ convergence $f(u_{n_k})\to f(u)$ follows by dominated convergence theorem. Since the limit is independent of the taken subsequence, $f(u_n)\to f(u)$ in $L^2$ follows.

  • $\begingroup$ Thanks for your answer. But here we can only prove the convergence of a subsequence not a whole sequence $\endgroup$ – whereamI Feb 6 '18 at 10:40
  • $\begingroup$ You can prove: every subsequence of $(f(u_n))$ contains another subsequence that converges to $f(u)$. This convergence principle proves the convergence of the whole sequence. $\endgroup$ – daw Feb 6 '18 at 10:46
  • $\begingroup$ Your statement is true. But I still feel such a argument kind of weird. Thank you very much! $\endgroup$ – whereamI Feb 6 '18 at 11:48

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