# Continuity of a Composition map in a finite measure space

I have a homework problem I am unable to finish. Could someone please help me proceed?

Problem: Suppose $(X, \mathcal{A}, \mu)$ is a finite measure space (that is, $\mu(x) < \infty$ ) and $f:\mathbb{R} \to \mathbb{R}$ is bounded and continuous. Define a composition mapping $F (u) = f \circ u$. Show that if $1\leq p < \infty$ , the map $F: L^p(X) \to L^p(X)$ is continuous.

My working: Let $(u_n)_{n\in \mathbb{N}}$ be a sequence of functions such that $\| u_n - u \|_p \to 0.$ We need to show that $\| F(u_n) - F(u) \|_p \to 0.$ Since $f$ is bounded, there exists $M\in \mathbb{R}$ such that $|f(x)| < M$ for all $x\in \mathbb{R}$. Thus a dominating function of $h_n(x) = | f( u_n(x) ) - f(u(x))|^p$ is $g(x) = (2M)^p$ and since $\int_X (2M)^p dx = (2M)^p \int_X d\mu = (2M)^p \mu(X) < \infty$, by the Dominated Convergence Theorem we have that $$\lim_{n\to\infty} \int_X | f( u_n(x) ) - f(u(x))|^p dx = \int_X \lim_{n\to\infty} | f( u_n(x) ) - f(u(x))|^p dx.$$ By the continuity of $f$, the absolute value function and exponentiation by $p$, we can then get $$\lim_{n\to\infty} \int_X | f( u_n(x) ) - f(u(x))|^p dx = \int_X | f( \lim_{n\to\infty} u_n(x) ) - f(u(x))|^p dx.$$ Now, from here I am unsure how to proceed. One thing I know is that if $(u_n)_{n\in \mathbb{N}}$ is a sequence of functions such that $\| u_n - u \|_p \to 0$, then there exists a subsequence $(u_{n_k})$ such that $u_{n_k}$ converges pointwise to $u$ almost everywhere. If we let $N$ denote the set where the pointwise convergence does not hold, then we have $$\int_X| f( \lim_{n\to\infty} u_{n_k}(x) ) - f(u(x))|^p dx = \int_{X/N} 0 dx + \int_N | f( \lim_{n\to\infty} u_{n_k} (x) ) - f(u(x))|^p dx =0$$ where the first integral is $0$ because the integrand is $0$, and the second integral is zero because $\mu(N) = 0$. So that shows the limit I want for at least a subsequence, but I don't know how to connect the others back to it. Any help would be greatly appreciated, but I would especially prefer if your answer finishes off my proof leaving most of it intact, rather than perhaps an entirely different start. Thank you.

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We can use the fact that a sequence which converges in $L^p$ has an almost-everywhere converging subsequence by the following way. Let $u\in L^p$ and assume that $F$ is not continuous at $u$. Then we can find a sequence $\{u_n\}\in L^p(X)$ and a $\delta >0$ such that $\lVert u-u_n\rVert_p$ converges to $0$ and $\lVert F(u)-F(u_n)\rVert_p\geq \delta$ for all $n$. Now, we can extract an almost-everywhere converging subsequence $\{u_{n_k}\}$. Since $F$ is continuous and bounded we can apply the dominated convergence theorem to get a contradiction. –  Davide Giraudo Oct 9 '11 at 17:09

## 1 Answer

We can use the fact that a sequence which converges in $L^p$ has an almost-everywhere converging subsequence by the following way. Let $u\in L^p$ and assume that $F$ is not continuous at $u$. Then we can find a sequence $\{u_n\}\in L^p(X)$ and a $\delta >0$ such that $$\lVert u-u_n\rVert_p \mbox{ converges to } 0 \mbox{ and }\lVert F(u)-F(u_n)\rVert_p\geq \delta\mbox{ for all }n.$$ Now, we can extract an almost-everywhere converging subsequence $\{u_{n_k}\}$. Since $F$ is continuous and bounded we can apply the dominated convergence theorem. Indeed, put $g_k\colon x\mapsto |f(u(x))-f(u_{n_k}(x))|^p$. Then $g_k\in L^1(X)$

• $|g_k(x)|\leq \left(2\sup_{t\in\mathbb R}|f(t)|\right)^p$ which is integrable since the measure of $X$ is finite.

• Let $N\in\mathcal A$ such that for all $x\in X\setminus N$ we have $\lim_{k\to +\infty}u_{n_k}(x)=u(x)$. The continuity of $f$ shows that we have $\lim_{k\to +\infty}f\circ u_{n_k}(x)=f\circ u(x)$ for all $x\in X\setminus N$.

We get that $\lim_{k\to \infty}\lVert F(u_{n_k})-F(u)\rVert_p=0$ whereas we have $\lVert F(u_{n_k})-F(u)\rVert_p\geq \delta$.

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(+1) Is it possible to make this into a direct proof? It feels like it could become one and then I could place this at the end of my current proof instead of restructuring it all. –  Mani Oct 9 '11 at 21:01
If you want a direct proof, you can use the following result: a sequence $\{x_n\}$ in a metric space converges to $x$ if and only if for all subsequence of $\{x_n\}$ we can extract from this subsequence a subsequence which converges to $x$. –  Davide Giraudo Oct 9 '11 at 21:29
Thank you for your help, I understand the proof now. –  Mani Oct 10 '11 at 3:13