Let $H^k(\mathbb{R}^n)$ be the Sobolev space $W^{k,2}(\mathbb{R}^n)$. Do we have $H^2(\mathbb{R}^n) \subset\ H^1(\mathbb{R}^n)$?

Question

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.

Answer 1

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.)

Answer 2

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)