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How do prove the inclusion of Sobolev spaces $H^2(\mathbb{R^2}) \subset W^{1,4}(\mathbb{R^2})$? Any ideas?

Ilovemath
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    I think is directly by definition.Take an element in $H^{2}(\mathbb{R}^{2})$ and then prove that the element is in $W^{1,4}(\mathbb{R}^{2})$. –  Dec 22 '20 at 02:23

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This is an immediate consequence of the general Sobolev inequality. This implies the embedding theorem, which says that if you have $k,l,p,q,n$ satisfying $$\frac{1}{p} - \frac{k}{n} = \frac{1}{q} - \frac{l}{n}$$

then you have $W^{k,p}(\mathbb{R}^n) \subseteq W^{l,q}(\mathbb{R}^n)$.

Since $H^2(X) = W^{2,2}(X)$ by definition, you can do the arithmetic and see that this holds.

A. Thomas Yerger
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  • I didn't know this version of Sobolev inequality, in which book can I find it? – Ilovemath Dec 22 '20 at 13:21
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    I confess I took this particular formulation straight from wikipedia. I learned the general theory of Sobolev spaces from Evans, but in the case that that does not treat this particular level of generality, I invite you to take a look at this MO thread, which will enlighten you on how to generalize what is in Evans to the desired statement(s): https://mathoverflow.net/questions/17736/way-to-memorize-relations-between-the-sobolev-spaces – A. Thomas Yerger Dec 27 '20 at 07:05