Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $||uv||_s \leq C ||u||_s ||v||_s$. Recall that $||f||_s=||(1+|\eta|^2)^{\frac{s}{2}}\hat{f}(\eta)||$, the norm on $H^s(\mathbb{R}^n)$. This is an exercise from Taylor's book, Partial differential equations I.

share|cite|improve this question
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
up vote 20 down vote accepted

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 $s>\frac{n}2$.

share|cite|improve this answer
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 3:45
@FrankZermelo: I updated the answer. – timur Mar 2 '13 at 1:45
Just beautiful! Thanks a lot! Very slick that you're not using Peetre's inequality with product, but with sum. – Frank Zermelo Mar 2 '13 at 2:22
@ timur, why does: $\| \hat{u} \| _{L^1} \leq C \| u \| _{H^s}$? – MathematicalPhysicist Dec 12 '14 at 7:28
@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 0:07

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 $H^{1/2+\varepsilon'}$.

share|cite|improve this answer
That's a pretty cute proof! +1 – Willie Wong Feb 26 '13 at 17:30
@WillieWong .. Thanks! :) – paul garrett Feb 26 '13 at 17:44
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
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 (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 0:10
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 at 14:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.