Preservation of relative compactness in different $L^p$ spaces.

Question

Let $\Omega$ be a measurable set $\mathcal{R}^{n}$ of finite measure. Show that if $\mathcal{H}$ is relatively compact in $L^{\infty}(\Omega)$ then $\mathcal{H}$ is also relatively compact in $L^{p}(\Omega)$.

A subset $H$ of metric space is said to be relatively compact if it has compact closure in $M$.

My thoughts: Since compact sets in metric spaces are equivalent to be complete and totally bounded my doubt is about if the closures are different with respect to these norms. It seems reasonable.

And if $\Omega$ is of finite measure, the $L^p$ norm is bounded by the $L^\infty$ norm.

Answer 1

There are various ways to prove it. Since $\Omega$ has finite measure, we have a continuous inclusion $\iota \colon L^\infty(\Omega) \to L^p(\Omega)$.

Way 1: Since by assumption $\mathcal{H}$ is relatively compact in $L^\infty(\Omega)$, the set $\iota(\overline{\mathcal{H}})$ is a compact subset of $L^p(\Omega)$ by the continuity of $\iota$, hence $\overline{\iota(\mathcal{H})} = \iota(\overline{\mathcal{H}})$ is compact in $L^p(\Omega)$.

Way 2: Since the $L^p$ spaces are complete, a subset is relatively compact if and only if it is totally bounded. By assumption, $\mathcal{H}$ is totally bounded, and since $\iota$ is a continuous linear map between normed spaces, it is Lipschitz continuous. Given $\varepsilon > 0$, there are finitely many $f_1,\dotsc, f_n \in \mathcal{H}$ such that

$$\mathcal{H} \subset \bigcup_{k = 1}^n B_{\varepsilon/\lVert\iota\rVert}^{L^\infty}(f_k).$$

By Lipschitz continuity, that implies

$$\iota(\mathcal{H}) \subset \bigcup_{k = 1}^n \iota\bigl(B_{\varepsilon/\lVert\iota\rVert}^{L^\infty}(f_k)\bigr) \subset \bigcup_{k = 1}^n B_\varepsilon^{L^p}(\iota(f_k)).$$

Since $\varepsilon > 0$ was arbitrary, this shows the total boundedness of $\iota(\mathcal{H})$.

Note that for sets of finite measure, both arguments work without modification for the continuous inclusion $L^q(\Omega) \hookrightarrow L^p(\Omega)$ whenever $1 \leqslant p \leqslant q \leqslant \infty$.

Answer 2

If $H\subset X,$ where $X$ is a metric space, then $H$ is relatively compact in $X$ iff every sequence in $H$ has a subsequence that converges in $X.$

Next note that because $\Omega$ has finite measure, $L^\infty(\Omega)\subset L^p(\Omega).$ Furthermore, convergence in $L^\infty(\Omega)$ implies convergence in $\subset L^p(\Omega).$ (Use Holder, or even more elementary means.)

So suppose $H$ is relatively compact in $L^\infty(\Omega).$ Let $f_n$ be a sequence in $H.$ Then there is a subsequence $f_{n_k}$ that converges in $L^\infty(\Omega).$ This implies $f_{n_k}$ converges in $L^p(\Omega).$ It follows that $H$ is relatively compact in $L^p(\Omega).$