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$(X, \mathcal{B}, \mu)$ is a measure space.

  • Is there any well-known criteria for compactness of a closed set in $L^2(X, \mu)$?

  • If the answer is negative what about $L^2(\mathbb{R}^n,\mu)$(in this part $\mu$ is Lebesgue measure)?

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1 Answer 1

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For Lebesgue measure, look at the Fréchet-Kolmogorov theorem.

For a general measure space, see Theorem 4.7.28 of Bogachev's Measure Theory, which is attributed to Riesz. Let $\pi = \{E_1, \dots, E_k\}$ be a collection of disjoint measurable subsets of $X$ having finite positive measure. ($\pi$ is almost a partition of $X$ but it can omit a set of cofinite measure.) For $f \in L^p(X,\mu)$, let $$\mathbb{E}^\pi f(x) = \begin{cases} \frac{1}{\mu(E_i)} \int_{E_i} f\,d\mu, & x \in E_i \\ 0, & x \notin \bigcup E_i.\end{cases}$$ Note that all the integrals exist since they are taken over sets of finite measure.

Write $\pi_1 \le \pi_2$ if every set in $\pi_1$ is a union of sets from $\pi_2$, up to sets of measure zero. Think of $\pi_2$ as being a sort of "refinement" of $\pi_1$, except that $\pi_2$ can also pick up some of $X \setminus \bigcup \pi_1$.

Then the result is that $K \subset L^p(\mu)$ is compact iff it is closed and bounded and $\mathbb{E}^\pi f \to f$ in $L^p$ norm, uniformly on $K$, as $\pi$ becomes increasingly refined. More precisely, a closed bounded $K$ is compact iff for every $\epsilon$ there exists $\pi_0$ such that for all $\pi \ge \pi_0$ and all $f \in K$ we have $\|\mathbb{E}^\pi f - f\|_{L^p(\mu)} < \epsilon$.

(Bogachev states it in the following equivalent way: view $\{\sup_{f \in K} \|\mathbb{E}^\pi f - f\|_{L^p}\}$ as a net in $[0,\infty)$ indexed by the set of all $\pi$. Then the necessary and sufficient condition is that this net should converge to 0.)

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  • $\begingroup$ What is the answer in general case when $X$ is an arbitrary space? @Nate Eldredge $\endgroup$ Mar 29, 2015 at 17:36
  • $\begingroup$ @GhiasiM: See my edit. $\endgroup$ Mar 29, 2015 at 18:02

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