Compactness of a special kind of Integral operators

Let $ (S(t))_{t>0}$ be a continuous operator from $ L^2(0,1)$ to its self and Let $ K$ be the operator $ $ \eqalign{ & K:{L^2}(0,1) \to {L^2}(0,1) \cr & f: \to (Kf)(x) = \int\limits_0^1 {k(s,x)S(s)f(s)ds} \cr} $ $ where $ $ k \in {L^2}(0,1) \times {L^2}(0,1)$ $ It is well known that if $ S=I$ then $ K$ is compact operator. What can I say about the compactness of $ K$ is this case? Thank you.

Compactness via closed sets

Do you know why compactness of a set, in a metric space, can’t be expressed in terms of closed sets?

That is, the next statement is false: Let $ (X,d)$ a metric space and $ A\subseteq X$ such that every closed cover of $ A$ has a finite subcover. Then $ A$ is compact (in the usual sense).

I’m trying to find a non compact set $ A$ on a metric space $ X$ such that, whenever $ \{F_\lambda\}$ is a collection of closed sets such that $ A\subseteq\cup_{\lambda}F_\lambda$ , there are $ \lambda_1,…,\lambda_n$ with $ A\subseteq\cup_{i=1}^n F_{\lambda_i}$ .

Compactness when mapping into a higher $L^p$ space and then back

Let $ q>p$ and let $ T:L^p[0,1] \to L^q[0,1]$ be a bounded linear operator. Let $ i: L^q[0,1] \to L^p[0,1]$ be the inclusion map (which is bounded). Is the composition $ i\circ T$ necessarily a compact operator on $ L^p$ ? We can assume $ p\in (1,\infty)$ if the uniform convexity of the underlying space helps in some way, though the boundary cases might be interesting themselves.

Context: This was inspired by this question, where the accepted answer covers the case $ q=\infty$ (in this case the composition is indeed always compact).

I’ve been at it for a couple days trying to prove and also to disprove it. It might be trivial with a silly oversight on my part, but on the other hand it could be inherently nontrivial as well, and I was hoping to learn some new fun fact about the geometry of Banach spaces in case of the latter.

Ideas: To prove it here are a couple ideas that I played around with. One suggestion is to somehow translate the problem to a statement about $ \ell^p$ spaces and then apply Pitt’s theorem. Unfortunately there seems to be no clear way to do this, for example any attempt at using a Fourier transform (which boundedly sends $ L^p \to \ell^{\frac{p}{p-1}}$ and $ \ell^p \to L^{\frac{p}{p-1}}$ for $ 1< p < 2$ ) will somehow “go the wrong way.” A second idea is to try to find a Banach space $ X$ which has the Dunford-Pettis property (e.g. a $ L^1(\mu)$ or $ C(K)$ space) and which nests in between $ L^p$ and $ L^q$ , i.e., $ L^q \subset X \subset L^p$ . This would easily show compactness of the inclusion, but finding such $ X$ seems not too easy. A third idea is to try to mimic the proof of Pitt’s theorem adapted to this $ L^p$ context. Basically suppose $ i\circ T$ was not compact. Then we can find $ f_n \in L^p$ with $ \|f_n\|_p=1$ and $ f_n \to 0$ weakly, and also $ \|Tf_n\|_p \ge \delta>0$ . Then (after perhaps passing to a subsequence) one may try to show that $ T$ is bounded below on the closed linear span of the $ f_n$ , which would imply that the image of $ T$ contains a closed infinite-dimensional subspace of $ L^p$ . And I’m farily certain that such a subspace cannot be contained in $ L^q$ . But formalizing these ideas might take some work.

On the other hand, here are some ideas for potential counterexamples if it turns out to be false. For one idea let us identify $ L^p[0,1] \simeq L^p(\Bbb R)$ and consider the Fourier transform $ \mathcal F:L^p(\Bbb R) \to L^{\frac{p}{p-1}}(\Bbb R)$ , where $ p<2$ . I’ve shown that $ \mathcal F$ is not a compact operator. Hence there is at least a noncompact operator from $ L^p[0,1] \to L^{\frac{p}{p-1}}[0,1]$ for $ p<2$ (which brings up another question of what if we restrict attention to $ q>p^*$ , then can we say that $ T$ itself is compact?… but perhaps that’s for another day). However, the difficult thing is to show that it remains noncompact when we compose it with the inclusion map, and I think that it actually becomes false. I played around with the Hermite functions $ \psi_n$ which orthonormally diagonalize $ \mathcal F$ on $ L^2(\Bbb R)$ , and I was able to compute that $ \|\psi_n\|_{L^p(\Bbb R)} \sim_n C_p n^{\frac1{3p}-\frac16}$ , so while these converge weakly to $ 0$ in $ L^2(\Bbb R)$ they fail to do so in $ L^p(\Bbb R)$ for $ p<2$ , hence they are useless for us. A second idea is to use some of the abstract mappings mentioned in this MO thread and the subsequent comments. Actually this might all be trivial from some proposition in Albiac-Kalton but I can’t access the book at the moment.

Compactness of the automorphic quotient and genericity

Let $ G$ be a reductive group defined over a field $ F$ . Let $ \mathbf{A}$ denote the ring of adeles of $ F$ . My question is:

Assuming the automorphic quotient $ [G]=G(F) \backslash G(\mathbf{A})$ is compact, can we say that all the (non-character) automorphic representations of $ G$ are tempered? generic?

I do not precisely understand the relations between the notions of tempered and generic representations beyond the very specific $ \mathrm{GL}(n)$ case, so that any reference about these matters is also welcome.

Metric space compactness

Consider the space $ C([0,1])$ equipped with the uniform norm. Find a sequence of functions $ \{g_n\}$ in $ C([0,1])$ so that $ \overline{\{g_n\}}$ is compact, but $ g_n$ does not converge uniformly.

I’m confused about $ \overline{\{g_n\}}$ and how to prove its compactness

Thanks!

Why is compactness important to find solution of certain problem?

I’ve been recently re-thinking about my knowledge of compacts sets and I’ve realized that other than the definition and maybe understanding with some struggle some proofs I don’t actually see why they’re very important. From some questions in this website which I used as sort of “soft discussion” about the subject it is my understanding that compactness might allow to state if certain functions are bounded or not.

However I came across this question, and I was considering the second answer specifically (more specifically the bit below):

Unlike their Holder cousins, most Sobolev spaces are reflexive Banach spaces. Reflexivity is a highly desirable feature for variational problems because it gives a little bit of compactness enough so you can prove the existence of minimizers (or more general critical points) of various energy functionals.

So this made me wonder… Is compactness in general a very highly desirable property that allow to prove existence of solutions of problems in general? If yes, how? I mean what is the general argument that is brought up.

how does knowing that some sets are compacts actually help in proving existence of solutions?