33

How do you prove that the Sobolev space $H^s(\mathbb{R}^n)$ is an algebra if $s>\frac{n}{2}$, i.e. if $u,v$ are in $H^s(\mathbb{R}^n)$, then so is $uv$? Actually I think we should also have $\lVert uv\rVert_s \leq C \lVert u\rVert_s \lVert v\rVert_s$. Recall that $\lVert f\rVert_s=\lVert(1+|\eta|^2)^{s/2}\,\hat{f}(\eta)\rVert$, the norm on $H^s(\mathbb{R}^n)$. This is an exercise from Taylor's book, Partial differential equations I.

Dominic Wynter
  • 5,909
  • 1
  • 11
  • 42
Frank Zermelo
  • 1,177
  • 9
  • 18
  • So that would mean that if I have some functions in $\mathbb R^3$ that are twice weakly differentiable, then all products between those functions would also be twice weakly differentiable? Interesting. – Elmar Zander Feb 26 '13 at 13:23
  • I think a proof of this is also given as Theorem 4.39 in Adams R. A., Fournier J. J.: Sobolev Spaces. Some readers may find the proof there easier than the one from the answers below. – user505117 Oct 15 '21 at 12:11

2 Answers2

36

Note that $$ \begin{split} (1+|\xi|^2)^p &\leq (1+2|\xi-\eta|^2+2|\eta|^2)^p\\ &\leq 2^p(1+|\xi-\eta|^2+1+|\eta|^2)^p\\ &\leq c(1+|\xi-\eta|^2)^p + c(1+|\eta|^2)^p, \end{split} $$ for $p>0$, where $c=\max\{2^{p},2^{2p-1}\}$. Put $\langle\xi\rangle=\sqrt{1+|\xi|^2}$. Then we have $$ \begin{split} \langle\xi\rangle^s |\widehat{uv}(\xi)| &\leq \int \langle\xi\rangle^s |\hat{u}(\xi-\eta)\hat{v}(\eta)|\,\mathrm{d}\eta\\ &\leq c\int \langle\xi-\eta\rangle^s |\hat{u}(\xi-\eta)\hat{v}(\eta)|\,\mathrm{d}\eta + c\int \langle\eta\rangle^s |\hat{u}(\xi-\eta)\hat{v}(\eta)|\,\mathrm{d}\eta\\ &\leq c|\langle\cdot\rangle^s\hat u|*|\hat v| + c|\hat u|*|\langle\cdot\rangle^s\hat v|, \end{split} $$ which, in light of Young's inequality, implies $$ \|uv\|_{H^s} \leq c\|u\|_{H^s} \|\hat v\|_{L^1} + c\|\hat u\|_{L^1}\|v\|_{H^s}. $$ Finally, we note that $\|\hat u\|_{L^1}\leq C\,\|u\|_{H^s}$ when $s>\frac{n}2$.

timur
  • 15,926
  • 25
  • 48
  • I tried to use that inequality but I get just $||fg||_s \leq ||f||_s ||(1+|\eta|^2)^s \hat{g}(\eta)||_{L^1}$, which is good if $g \in C_{0}^{\infty}$, but not always on $H^s(\mathbb{R}^n)$. – Frank Zermelo Feb 27 '13 at 03:45
  • @FrankZermelo: I updated the answer. – timur Mar 02 '13 at 01:45
  • 3
    @ timur, why does: $\| \hat{u} \| _{L^1} \leq C \| u \| _{H^s}$? – MathematicalPhysicist Dec 12 '14 at 07:28
  • 4
    @MathematicalPhysicist: Write $\|\hat u\|_{L^1}=\int<\xi>^{-s}<\xi>^{s}|\hat u(\xi)|\mathrm{d}\xi$, and apply Cauchy-Schwarz. The condition $2s>n$ gives integrability of $<\xi>^{-2s}$. – timur Dec 14 '14 at 00:07
  • @ timur, thanks for the answer but I think you showed that $\int (1+|\xi|^2)^{s/2}\widehat{uv}(\xi)d\xi \leq c\|u\|_{H^s} \|\hat v\|_{L^1} + c\|\hat u\|_{L^1}\|v\|_{H^s}$, and $\int (1+|\xi|^2)^{s/2}\widehat{uv}(\xi)d\xi \neq \|uv\|_{H^s}$. Please help me. – David Alexander Dec 10 '21 at 00:50
  • @DavidAlexander: You take the L2-norm of that expression, not just integration – timur Dec 10 '21 at 02:34
  • @ timur, you are right, I am very tired. Thank you so much! – David Alexander Dec 10 '21 at 03:54
  • @timur: Don't the first set of inequalities also work for $p=0$ implying that $p\geq 0$? – Axion004 Jan 20 '22 at 07:52
6

One way to see this is by an argument similar to proof of a "trace theorem": first, for $f,g\in H^s(\mathbb R^n)$ with $s\ge 0$, $f\otimes g\in H^s(\mathbb R^{2n})$ because $1+|x|^2+|y|^2\le (1+|x|^2)(1+|y|^2)$. Next, prove an easy form of a "trace theorem", namely, that restriction from $\mathbb R^N$ to $\mathbb R^{N-n}$ maps $H^s$ to $H^{s-{n\over 2}}$ for $s> {n\over 2}$.

Edit: in response to the comments of @ElmarZander, ... The question, as originally posed, cannot be quite right, no. The argument sketched here shows that for $s>n/2$ the product of two elements in $H^s(\mathbb R^n)$ is in $H^{s-n/2-\varepsilon}$ for every $\epsilon>0$. I do not know whether higher-dimensional results can be sharpened, but for $n=1$ it is easy to do explicit examples showing the limitation: take $\hat{f}=\hat{g}$ to be $|x|^{-3/4-\varepsilon}$ for $x\ge 1$ and $0$ otherwise. These are in $H^{1/2+\varepsilon'}(\mathbb R)$. Then the convolution has a lower bound $x^{-1/2-2\varepsilon}$, I believe, so $fg$ is not in $H^{1/2+\varepsilon'}$.

Giuseppe Negro
  • 28,987
  • 6
  • 59
  • 199
paul garrett
  • 46,394
  • 4
  • 79
  • 149
  • Forgive my ignorance, but if I get you correctly you restrict $\mathbb R^{2n}$ to $D=\{(x,x)|x\in\mathbb R^{n}\}$ and then map $H^s(\mathbb R^{2n})$ to $H^{s-n/2}(D)\simeq H^{s-n/2}(\mathbb R^{n})$. But then you lose differentiability, don't you? – Elmar Zander Feb 26 '13 at 18:26
  • Excuse me for my confusion, but isn't the restriction giving you the bound of $uv$ in $H^{s-n/2}(\mathbb{R}^n)$, and not in $H^s(\mathbb{R}^n)$? Also, isn't it clear just from the fact that $||f||_s ||g||_s$ $\geq ||f \otimes g||_{H^s(\mathbb{R}^{2n})} \geq $ $||f \otimes g||_{H^s(D)} \geq ||fg||_{H^s(\mathbb{R}^{n})}$, where $D$ is the diagonal on $\mathbb{R}^{2n}$? Is it ok to consider the space $H^s(D)$, is the space $D$ sufficiently nice in order to consider the Sobolev space on it? I am sorry again if I do not see something obvious or already explained here. – Frank Zermelo Feb 27 '13 at 03:35
  • @FrankZermelo: Yes, the argument given does not prove what your question asks, but what it does prove is certainly true, and standard, etc. I think there is something amiss with the claim that multiplication can stay in the same Sobolev space, as the example on $\mathbb R^1$ shows. As to your specific questions: yes, it's easy to see that $f\otimes g$ is in the expected space; yes, the diagonal $\mathbb R^n$ is nice enough to look at the restriction mapping (the "trace theorem" instance). There is where something equivalent to the "Peetre inequality" arises. – paul garrett Feb 27 '13 at 13:50
  • 2
    For $s>n/2$, we have the a priori estimate $\|fg\|_{W^{s,p}}\lesssim_{n,p,s}\|f\|_{W^{s,p}}\|g\|_{W^{s,p}}$. See T. Tao's [lecture notes](http://www.math.ucla.edu/~tao/254a.1.01w/) (Week 4) for a proof. $H^{s}(\mathbb{R}^{n})=W^{s,2}(\mathbb{R}^{n})$. So this contradicts your claim. – Matt Rosenzweig Dec 26 '15 at 00:10
  • 1
    If $f,g\in H^1(\mathbb{R})$, then an application of Leibniz' rule, together with $\|\cdot\|_{L^\infty}\leq \|\cdot\|_{H^1(\mathbb{R})}$ and a limiting argument, shows that $fg\in H^1(\mathbb{R})$. I don't see why this shouldn't work for any integer-regularity Sobolev space that controls $L^\infty$ – Bananach Jan 28 '16 at 14:45