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Starting on the subject of Sobolev spaces I got the doubt: Is $H^2(\mathbb{R}^n) \subset\ H^1(\mathbb{R}^n)$? I think it's clearly right because of by the definition of the norm in $H^k(\mathbb{R}^n)$ spaces we have that functions in $H^2(\mathbb{R}^n)$ must be in $L^2(\mathbb{R}^n)$ as well as their first and second derivatives. So, such a function is obviously in $H^1(\mathbb{R}^n)$, right? If this is a correct guess that's ok! But my doubt is because I didn't found up to now any general theorem supporting that statement, e.g., embedding Sobolev ones. Thanks.

R. Ferreira
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2 Answers2

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Yes. Indeed, for $s>t$, $H^s(\mathbb{R}^n) \subset H^{t}(\mathbb{R}^n)$. The norm may be expressed in terms of the Fourier transform as $$ \lVert u \rVert_{H^s}^2 = \int_{\mathbb{R}^n} (1+\lvert k\rvert^2)^{s/2} \lvert \hat{u}(k) \rvert^2 \, dk, $$ and of course $ (1+\lvert k\rvert^2)^{s/2} = (1+\lvert k\rvert^2)^{t/2} (1+\lvert k\rvert^2)^{(s-t)/2} $, and the latter factor is at least $1$, so $$ \lVert u \rVert_{H^s}^2 = \int_{\mathbb{R}^n} (1+\lvert k\rvert^2)^{t/2} (1+\lvert k\rvert^2)^{(s-t)/2} \lvert \hat{u}(k) \rvert^2 \, dk \geq \int_{\mathbb{R}^n} (1+\lvert k\rvert^2)^{t/2} \lvert \hat{u}(k) \rvert^2 \, dk =\lVert u \rVert_{H^t}^2, $$ so if the $H^s$-norm is bounded, so is the $H^t$-norm. One can work with the norm in real space in a similar way, but this is an easy way to think in $\mathbb{R}^n$, and applies to any $s$, not just integers.

(That the containment is strict is easily dealt with by constructing a Fourier transform with decay qualities that are good enough for $H^t$ but not $H^s$, in the usual way.)

Chappers
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  • Thank you very much Chappers! Perfect answer! Solved my problem. Do you think that this can be seen by means of some existing general theorem? I mean very known embedding theorems. – R. Ferreira May 05 '17 at 00:01
  • Phrased like this, it's just saying that if $0 < g \leq f$, $L^2(\mathbb{R}^n, f(k) , dk) \subset L^2(\mathbb{R}^n, g(k) , dk) $. I don't think it can be made any simpler than that. – Chappers May 05 '17 at 02:03
  • (By the way, it's good etiquette around here to upvote an answer you find helpful, and to accept an answer if you feel it is the best answer to your question. If and when to accept an answer is up to you, and of course depends on the quality of the answers you receive.) – Chappers May 05 '17 at 02:06
  • Can you see the green see the green symbol of acceptance? I'm asking that for you because the green in my symbol seems to disappear. By the way, I have marked that. – R. Ferreira May 05 '17 at 17:09
  • Yes thank you, that's fine. – Chappers May 05 '17 at 17:16
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Of course true, since the two spaces are defined on L^2(R^n), then we can conclude the following recursion relation (see image/clic here)

MedSI
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  • Thanks for your answer. A different way, but an answer. I marked your text as useful, but I still don't understand why is the number zero between the up and down arrows. – R. Ferreira May 05 '17 at 17:14