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.
 5,909
 1
 11
 42
 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 Answers
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 $c=\max\{2^{p},2^{2p1}\}$. 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 $s>\frac{n}2$.
 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


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 CauchySchwarz. 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 L2norm of that expression, not just integration – timur Dec 10 '21 at 02:34


@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
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 $\mathbb R^{Nn}$ 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 $H^{sn/2\varepsilon}$ for every $\epsilon>0$. I do not know whether higherdimensional 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 $x^{1/22\varepsilon}$, I believe, so $fg$ is not in $H^{1/2+\varepsilon'}$.
 28,987
 6
 59
 199
 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 $H^{sn/2}(D)\simeq H^{sn/2}(\mathbb R^{n})$. 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 $H^{sn/2}(\mathbb{R}^n)$, 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

2For $s>n/2$, 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

1If $f,g\in H^1(\mathbb{R})$, 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 integerregularity Sobolev space that controls $L^\infty$ – Bananach Jan 28 '16 at 14:45