How is the dominated convergence theorem applied here?

Question

Let $(E,\mathcal E,\mu)$ be a probaiblity space and $A_n,B$ be linear contractions (operator norm at most $1$) on $L^p(\mu)$ for all $p\in[1,\infty]$. Say we know that $$\left\|(A_n-B)f\right\|_{L^2}\xrightarrow{n\to\infty}0\tag1\;\;\;\text{for all }f\in\mathcal L^2(\mu).$$

I need help to understand the following argument which aims to conclude $$\left\|(A_n-B)f\right\|_{L^p}\xrightarrow{n\to\infty}0\tag2\;\;\;\text{for all }f\in\mathcal L^p(\mu)$$ for all $p\in[1,\infty]$ from $(1)$:

If $f\in\mathcal L^\infty(\mu)$, then (since $A_n$ is a contraction) $(A_nf)_{n\in\mathbb N}$ is bounded in $L^\infty(\mu)$ which means that it is uniformly bounded $\mu$-almost surely. Now I've read that $\left\|(A_n-B)f\right\|_{L^p}\to0$ follows from $(1)$ and the dominated convergence theorem.

I don't get that. Clearly, $(A_nf)_{n\in\mathbb N}$ is uniformly bounded a.s. and hence trivially dominated by a $L^p$-integrable function for all $p\in[1,\infty]$. However, I don't get how we can utilize $(1)$ now. For the dominated convergence theorem we would need a.s. pointwise convergence of $A_nf$ to $Bf$ (so, for example, $A_nf\to Bf$ in $L^\infty$) ...

(If this could be shown, then the extension to all $f\in L^p$ would be easy, since the elementary functions (which are in $L^\infty$) are dense in $L^p$ for all $p\in[1,\infty]$.)

Answer 1

This result is only true for $p \in [1, \infty)$. Let me start by giving a counterexample for the case $p = \infty$. Let $E = [0,1]$ with its Borel $\sigma$-algebra and the Lebesgue measure. Let $A_n f := 1_{[0, 1-\frac1n]} f$ and $B = \operatorname{Id}$. It is easy to check that these are both contractions in $L^q$ for every $q \in [1,\infty]$ and to check that $$\|A_n f - Bf \|_{L^p} \to 0$$ for every $p \in [1, \infty)$, say by using the D.C.T. However if $f(x) = 1$ for all $x \in [0,1]$ then $\|A_n f - B f\|_\infty = 1$ for all $n$.

A proof for the case $p \in [1,\infty)$ was outlined in my comment, which I recreate here. First let $I_n = A_n f - B f$. By contractivity of $A_n,B$ in $L^\infty$, we can find an $L^\infty$ function $g$ such that $|I_n| \leq g$.

Now consider an arbitrary subsequence $I_{n_k}$. By the convergence in $L^2$, this subsequence has a further subsequence that converges a.e. Then, for $p \in [1,\infty)$, by the DCT with dominating function $|g|^p$, that further subsequence converges to $0$ in $L^p$.

The desired convergence then follows by a standard subsequences argument; a sequence $x_n$ in a topological space converges to $x$ if and only if every subsequence of $x_n$ has a further subsequence converging to $x$.

Answer 2

The answer given by Rhys Steele is of course perfectly fine. Here is an alternative argument: for $1\leqslant p\lt 2$, this follows from the fact that $\lVert g\rVert_p\leqslant\lVert g\rVert_2$ for all $g\in\mathbb L^2$. For $p>2$, write $$ \left\|(A_n-B)f\right\|_{L^p}^p=\mathbb E\left[\lvert A_nf-Bf\rvert^p\right] \leqslant\mathbb E\left[\lvert A_nf-Bf\rvert^2\right]\lVert A_nf-Bf\rVert_\infty^{p-2} $$ and $\lVert A_nf-Bf\rVert_\infty\leqslant 2\lVert f\rVert_\infty$ hence $$ \left\|(A_n-B)f\right\|_{L^p}^p\leqslant 2^{p-2}\lVert f\rVert_\infty^{p-2}\left\|(A_n-B)f\right\|_{L^2}^2. $$