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Can you help me to explain in detail why we deduce from Theorem 8.8 that the injection of $H^1(I)$ into $L^2(I)$ is compact.

enter image description here

I understand that

  1. $H^1(I)$ is compact embedded in $C(\bar I)$, and
  2. $C(\bar I)$ is continuously embedded in $L^2(I)$

However, I can not see that "the injection of $H^1(I)$ into$L^2(I)$ is compact" follows from 1. and 2.

Rohit Singh
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Long Thanh
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    Welcome to MSE. It is in your best interest that you type your posts (using MathJax) instead of posting links to pictures. – José Carlos Santos Feb 28 '22 at 08:58
  • Hint: $C(\bar I)$ is continuously embedded in $L^2(I)$ when $I$ is bounded. Does this help? – Claudio Moneo Feb 28 '22 at 10:46
  • Thank you @ClaudioMoneo. I understand that $H^1(I)$ is compact embedded in $C(\overline{I})$ (1) and $C(\overline{I})$ is continuously embedded in $L^2(I)$ (2). However, I can not see that "the injection of $H^1(I)$ into $L^2(I)$ is compact" follows from (1) and (2). – Long Thanh Feb 28 '22 at 15:53

3 Answers3

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Using the Cauchy-Schwarz inequality, one get that a function in $H^1[a,b]$ satisfies $\vert f(x)-f(y)\vert \leq \vert \vert f\vert \vert _{H^1} \sqrt {x-y}$.

Therefore, the image of the unit ball is a family of uniformly equicontinuous (in fact 1/2Hölder) functions.

In order to use Arzela-Ascoli's theorem, it remain to prove that this image is bounded. But if $f\in H^1[a,b], \vert {1\over b-a}\int _a^bf(t) dt\vert \leq C \vert \vert f\vert \vert _{L^2} \leq C \vert \vert f\vert \vert _{H^1}\leq C$, so that the mean of $f$ is bounded.

Then as $\vert f(x)-f(x_0)\vert \leq \vert \vert f\vert \vert _{H^1} \sqrt {x-x_0}$, choosing $x_0$ so that $f(x_0$ is the mean value of $f$, we see that $\vert f(x) \vert \leq \vert \vert f\vert \vert _{H^1} \sqrt {b-a}+C$, and the result follows.

Thomas
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The following is a simple but useful fact: Given a chain of continuous embeddings on Banach spaces $A \subset B, B\subset C, ..., C \subset N$, it suffices for only one of all the embeddings to be compact for the whole chain to be compact.

This can be verified by using the definition of sequential compactness (given a bounded sequence in $A$, choose a convergent subsequence).

In your particular case, all we now have to see is that

  • $H^1(I)$ is compactly embedded in $C(\bar I)$
  • $C(\bar I)$ is continuously embedded in $L^2(I)$

Now the first part follows immediately by the theorem you linked.

The second one is also easy too see if you consider that every $f \in C(\bar I)$ is bounded and thus square integrable.

Claudio Moneo
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A completely different proof is to use Fourier series. We may assume $I=[-/pi,\pi]$. let $e_n, n\in \bf Z$ the standard basis of $L^2(I)$, $e_n(t)= {1\over \sqrt 2\pi} e^{int}$. Certainly $f_n= {e\over \sqrt {n^2+1}}$ is an orthonormal basis of $H^1$, so that the natural embedding in this basis has a diagonal matrix with eigenvalues $ 1\over \sqrt {n^2+1}$. From this we get that if $P_n$ is the projector onto the subspace generated by $e_k, \vert k \vert \leq n$, $I$ the stabdard embedding and $I_n= I\circ P_n$, $\vert \vert I- I_n \vert \vert \leq {1\over \sqrt {(n+1)^2+1}}$, so that $I$ is as close as you want from a finite range operator

Thomas
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