$W^{s,p}(\mathbb{R}^{n})$ Is Not Closed Under Multiplication when $s\leq n/p$

Question

For $s\in\mathbb{R}$, $1<p<\infty$, and $n\geq 1$, define the Sobolev space $W^{s,p}(\mathbb{R}^{n})$ by $$W^{s,p}(\mathbb{R}^{n}):=\left\{f\in\mathcal{S}(\mathbb{R}^{n}) : \|(\langle{\xi}\rangle^{s}\widehat{f})^{\vee}\|_{L^{p}}<\infty\right\}$$, equipped with norm $$\|f\|_{W^{s,p}}=\|(\langle{\xi}\rangle^{s}\widehat{f})^{\vee}\|_{L^{p}}$$ If $\phi+\sum_{k=1}^{\infty}\psi(2^{-k}\cdot)=1$ is a Littlewood-Paley partition of unity, then we have the equivalence of norms $$\|f\|_{W^{s,p}}\approx \|P_{\leq 0}f\|_{L^{p}}+\left\|\left(\sum_{k=1}^{\infty}2^{2ks}|P_{k}f|^{2}\right)^{1/2}\right\|_{L^{p}},$$ where $\widehat{P_{\leq 0}f}=\psi\widehat{f}$ and $\widehat{P_{k}f}=\psi(2^{-k})\widehat{f}$.

When $sp> n$, one can show that $W^{s,p}$ is closed under pointwise multiplication. More precisely,

Theorem. For $n\geq 1$, $1<p<\infty$, and $s>n/p$, $$\|fg\|_{W^{s,p}}\lesssim_{n,p,s}\|f\|_{W^{s,p}}\|g\|_{W^{s,p}},\quad\forall f,g\in W^{s,p}(\mathbb{R}^{n})$$

I want to show that this result is sharp for any $1<p<\infty$, $n\geq 1$, and $0<s\leq n/p$. Specifically,

Problem. For all $n\geq 1$, $0<s\leq n/p$, find a function $f\in W^{s,p}(\mathbb{R}^{n})$ such that $f^{2}\notin W^{s,p}(\mathbb{R}^{n})$.

The proof of the above theorem with which I am familiar (see Week 4 notes here) uses the fact that by Sobolev embedding, we can control $\|f\|_{L^{\infty}}$ by $\|f\|_{W^{s,p}}$ and therefore we can control the $L^{\infty}$-norm of the Hardy-Littlewood maximal function of $f$, denoted $Mf$, by $\|f\|_{W^{s,p}}$. I believe for $s<n/p$, I can construct a function $f\in W^{s,p}$ which is not bounded. T. Tao writes that one cannot control the $L^{\infty}$ norm by the $W^{s,p}$ norm also when $s=n/p$, but I couldn't show the endpoint case (Edit 1: See below for proof of endpoint case).

Consider functions of the function $$f_{N}:=N^{-1}\sum_{k=1}^{N}2^{-sk}2^{nk/p}\widehat{\psi}(2^{k}x)$$ where $\psi$ is a nonnegative bump function adapted to the annulus $1\leq|\xi|\leq 2$ and $N\geq 1$ is an integer. Instead of $N^{-1}$, Tao uses $N^{-1/p}$ but I was unable to make that work.

I claim that $\|f_{N}\|_{W^{s,p}}\lesssim 1$ (i.e. uniformly in $N$). It's clear from the triangle inequality and dilation invariance that $\|f_{N}\|_{L^{p}}\lesssim 1$. Whence by Young's inequality, $\|P_{\leq 0}f_{N}\|_{L^{p}}\lesssim 1$. By analyzing frequency supports, we see that $${P_{k}2^{jn/p}\widehat{\psi}(2^{j}\cdot)}=0,\quad |k-j|> 3$$ Whence, $$2^{sk}P_{k}f_{N}=N^{-1}\sum_{j=k-3}^{k+3}P_{k}2^{s(k-j)}2^{jn/p}\widehat{\psi}(2^{j})=N^{-1}\sum_{j=-3}^{3}2^{-sj}P_{k}2^{(k+j)n/p}\widehat{\psi}(2^{j+k})$$ where terms with negative indices are defined to be zero. Thus by nesting of $\ell^{p}$ spaces and the above observations, we have that \begin{align*} \left(\sum_{k=1}^{\infty}|2^{sk}P_{k}f_{N}|^{2}\right)^{1/2}\lesssim_{n,s,p} N^{-1}\sum_{j=1}^{N}\sum_{k=j-3}^{j+3}|P_{k}2^{nj/p}\widehat{\psi}(2^{j})| \end{align*} By triangle inequality, Young's inequality, dilation invariance, we obtain that $$\left\|N^{-1}\sum_{j=1}^{N}\sum_{k=j-3}^{j+3}|P_{k}2^{nj/p}\widehat{\psi}(2^{j})|\right\|_{L^{p}}\lesssim_{n,s}N^{-1}\sum_{j=1}^{N}\|2^{nj/p}\widehat{\psi}(2^{j})\|_{L^{p}}=\|\widehat{\psi}\|_{L^{p}}$$

But since $\widehat{\psi}(0)=\int\psi dx=c>0$, it follows that $$\|f_{N}\|_{L^{\infty}}\gtrsim N^{-1}\sum_{k=1}^{N}2^{k(n/p-s)}\rightarrow\infty,$$ as $N\rightarrow\infty$, since $s<n/p$. By taking an increasing sequence $N_{k}\rightarrow\infty$ such that $\|f_{N_{k}}\|_{L^{\infty}}$ is suitably large, one can probaby construct a $W^{s,p}$ function of the form $f=\sum_{k}2^{-k}f_{N_{k}}$ such that $\|f\|_{L^{\infty}}=\infty$, but I don't see how to apply this to show what I want.

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Edit 1: I believe I have figured out how to show the endpoint case $s=n/p$ in the claim that $\|\cdot\|_{W^{s,p}}$ does not control $\|\cdot\|_{L^{\infty}}$.

Consider functions of the function $$f_{N}:=\sum_{k=1}^{N}2^{-sk}2^{nk/p}\widehat{\psi}(2^{k}x)$$ where $\psi$ is a nonnegative bump function adapted to the annulus $1\leq|\xi|\leq 2$ and $N\geq 1$ is an integer.

I claim that $\|f_{N}\|_{W^{s,p}}\lesssim_{n,s,p}N^{1/p}$ (i.e. uniformly in $N$). It's clear from the triangle inequality and dilation invariance that $\|f_{N}\|_{L^{p}}\lesssim N^{1/p}$. Whence by Young's inequality, $\|P_{\leq 0}f_{N}\|_{L^{p}}\lesssim N^{1/p}$. By analyzing frequency supports, we see that $${P_{k}2^{jn/p}\widehat{\psi}(2^{j}\cdot)}=0,\quad |k-j|> 3$$ Whence, $$2^{sk}P_{k}f_{N}=\sum_{j=k-3}^{k+3}P_{k}2^{s(k-j)}2^{jn/p}\widehat{\psi}(2^{j})=\sum_{j=-3}^{3}2^{-sj}P_{k}2^{(k+j)n/p}\widehat{\psi}(2^{j+k})$$ where terms with negative indices are defined to be zero. By the upper Littlewood-Paley inequality, \begin{align*} \left\|\left(\sum_{k=1}^{\infty}2^{2sk}|P_{k}f_{N}|^{2}\right)^{1/2}\right\|_{L^{p}}&\lesssim_{n,s,p}\left\|\left(\sum_{j=1}^{N}2^{2jn/p}|\widehat{\psi}(2^{j}\cdot)|^{2}\right)^{1/2}\right\|_{L^{p}}\\ &\leq\left\|\sum_{j=1}^{N}2^{jn/p}|\widehat{\psi}(2^{j}\cdot)|\right\|_{L^{p}} \end{align*} by the nesting property. Since $\widehat{\psi}$ is a Schwartz function adapted to a ball of radius $O(1)$, we have that $$\sum_{j=1}^{N}2^{jn/p}|\widehat{\psi}(2^{j}x)|\lesssim_{M}\sum_{j=1}^{N}\dfrac{1}{(1+|2^{j}x|)^{M}}$$ for any $M\geq 0$. So for integer $N-1\geq k\geq 0$, \begin{align*} \int_{2^{-k-1}\leq|x|\leq 2^{-k}}\left|\sum_{j=1}^{N}2^{jn/p}|\widehat{\psi}(2^{j}x)|\right|^{p}dx&\lesssim_{M}\int_{2^{-k-1}\leq|x|\leq 2^{-k}}\left|\sum_{j=1}^{k}2^{jn/p}+\sum_{j=k+1}^{N}\dfrac{2^{jn/p}}{(1+|2^{j}x|)^{M}}\right|^{p}dx\\ &\lesssim\int_{2^{-k}\leq|x|\leq 2^{-k-1}}\left|2^{kn/p}+\sum_{j=k+1}^{N}\dfrac{2^{jn/p}}{(1+|2^{j}x|)^{M}}\right|^{p}dx\\ \end{align*} The second term in the integrand is bounded from above by a decreasing, convergent geometric series, provided $M$ is sufficiently large, and is therefore comparable to its first term $\sim 2^{kn/p}$. Whence the above is majorized by \begin{align*} \lesssim\int_{2^{-k-1}\leq|x|\leq 2^{-k}}2^{kn}dx\sim_{n} 1 \end{align*} For $k\leq N$, we the estimate $$\int_{|x|\leq 2^{-N}}\left|\sum_{j=1}^{N}2^{jn/p}|\widehat{\psi}(2^{j}x)|\right|^{p}dx\lesssim\int_{|x|\leq 2^{-N}}2^{Nn}dx\sim_{n}1$$ For $|x|\geq 1$, rapid decay gives the estimate \begin{align*} \int_{|x|\geq 1}\left|\sum_{j=1}^{N}2^{jn/p}|\widehat{\psi}(2^{j}x)|\right|^{p}dx\lesssim_{M}\int_{|x|\geq 1}\left(\sum_{j=1}^{N}2^{jn/p}|2^{j}x|^{-M}\right)^{p}dx\lesssim_{n,M,p}1, \end{align*} provided $M$ is sufficiently large. Combining the estimates and adding up the pieces of the integral, we conclude that \begin{align*} \int_{\mathbb{R}^{n}}\left|\sum_{j=1}^{N}2^{jn/p}|\widehat{\psi}(2^{j}x)|\right|^{p}dx&\lesssim_{n,p} N \end{align*} Taking $p^{th}$ roots completes the proof of the claim.

Answer 1

Unless I'm missing something obvious, I believe I have a complete answer that demonstrates the existence of an $f\in W^{s,p}(\mathbb{R}^{n})$, for $0<s\leq n/p$ using a combination of hard analysis and soft analysis.

First, a small lemma:

Lemma 1. If $f\in W^{s,p}(\mathbb{R}^{n})$ is such that the operator defined by pointwise multiplication by $f$ is bounded on $W^{s,p}(\mathbb{R}^{n})$, then $f\in L^{\infty}(\mathbb{R}^{n})$.

Proof. Let $T$ denote the multiplier operator defined by $f$. By hypothesis $T$ is bounded $W^{s,p}\rightarrow W^{s,p}$. For all $g,h\in\mathcal{S}(\mathbb{R}^{n})$, we have that $$\int_{\mathbb{R}^{n}}T(g)\overline{h}=\int_{\mathbb{R}^{n}}fg\overline{h}=\int_{\mathbb{R}^{n}}g\overline{\overline{f}h}$$ By density, we see that the adjoint operator $T^{*}$ densely defined by multiplication by $\overline{f}$ is bounded $W^{-s,p'}\rightarrow W^{-s,p'}$, where $1/p+1/p'=1$. Taking complex conjugates, we see that the operator densely defined by multiplication by $f$ is bounded $W^{-s,p'}\rightarrow W^{-s,p'}$. Interpolating (see T. Tao's Notes 5 here), we see that multiplication by $f$ is bounded on $W^{s_{\theta},p_{\theta}}$, for all $$s_{\theta}:=\theta s-(1-\theta)s \quad p_{\theta}:=\dfrac{\theta}{p}+\dfrac{1-\theta}{p'}, \qquad\forall \enspace 0<\theta<1$$ In particular, for $\theta=1/2$, we obtain that the operator $T$ is bounded on $W^{0,2}=L^{2}$. Testing $T$ on simple functions, we see that $f$ is essentially bounded. $\Box$

Lemma 2. $W^{s,p}(\mathbb{R}^{n})\not\subset L^{\infty}(\mathbb{R}^{n})$, for $0<s\leq n/p$, $1<p<\infty$.

Assuming the lemma, take an $f\in W^{s,p}(\mathbb{R}^{n})\setminus L^{\infty}(\mathbb{R}^{n})$. By Lemma 1, there exists $g\in W^{s,p}(\mathbb{R}^{n})$ such that $fg\notin W^{s,p}(\mathbb{R}^{n})$. Define $h:=f+g\in W^{s,p}(\mathbb{R}^{n})$. If $f^{2}$ or $g^{2}\notin W^{s,p}(\mathbb{R}^{n})$, then we're done. Otherwise by the reverse triangle inequality, $$\|h^{2}\|_{W^{s,p}}\geq 2\|fg\|_{W^{s,p}}-\|f^{2}\|_{W^{s,p}}-\|g^{2}\|_{W^{s,p}}=\infty$$

We now prove Lemma 2. If $W^{s,p}$ embeds (not necessarily continuously) in $L^{\infty}$, then I claim it must do so boundedly. Indeed, Let $I$ denote the inclusion map. By the Closed Graph Theorem, it suffices to show that if $f_{n}\rightarrow f$ in $W^{s,p}$ and $If_{n}\rightarrow g$ in $L^{\infty}$, then $If=g$. Observe that $f_{n}\rightarrow f$ and $f_{n}\rightarrow g$ in distribution and therefore $f=g$ a.e. We arrive at a contradiction, as my analysis in the original post shows that that $W^{s,p}(\mathbb{R}^{n})$ does not continuously embed in $L^{\infty}(\mathbb{R}^{n})$.

Answer 2

What follows is not a solution to the original problem, but rather some length comments, which I thought best to post separately, so as to minimize the length of the original post.

First, it's not hard to see that for general $0<s\leq n/p$, it is insufficient to work at the $L^{p}$ level. Indeed, if $q$ is the Sobolev conjugate given by $1/q=1/p-s/n$ and $1/q\geq 1/2p$, then by Sobolev embedding we have that $\|f\|_{L^{2p}}\lesssim_{n,s,p}\|f\|_{W^{s,p}}$. In particular, this is true for the endpoint case $s=n/p$.

Second, it is sufficient to find an $f\in W^{s,p}(\mathbb{R}^{n})$ such that $$\|\sup_{k}|2^{ks}P_{k}f^{2}|\|_{L^{p}}=\infty$$ For those of you familiar with Besov spaces, this condition is equivalent to $\|f^{2}\|_{B_{p,\infty}^{s}}=\infty$. I mention this condition because I imagine that the $f\mapsto f^{2}$ is sharply bounded (i.e. the regularity parameter of the target space is sharp) from $W^{s,p}(\mathbb{R}^{n})\rightarrow W^{s-\epsilon,p}(\mathbb{R}^{n})$, for some $\epsilon>0$, and therefore we have the inclusions $$W^{s-\epsilon,p}(\mathbb{R}^{n})\subset B_{p,\infty}^{s-\delta}(\mathbb{R}^{n})\subset W^{s,p}(\mathbb{R}^{n}),\quad\forall \enspace 0<\delta<\epsilon$$

Third, when $0<s<n/p$, it is not too difficult to show that the nonlinear operator $f\mapsto f^{2}$ is unbounded on $W^{s,p}(\mathbb{R}^{n})$. Indeed, let $\psi$ be a nonnegative bump function adapted to the annulus $1\leq|\xi|\leq 2$ and consider $f_{j}=2^{-sj+nj/p}\widehat{\psi}(2^{j}\cdot)$, for positive integer $j$. Our analysis above shows that $\|f_{j}\|_{W^{s,p}}\lesssim_{n,s,p}1$. By dilation invariance, \begin{align*} \|2^{sk}P_{k}f_{k}^{2}\|_{L^{p}}^{p}&=2^{skp}\int_{\mathbb{R}^{n}}\left|\int_{\mathbb{R}^{n}}2^{kn}\varphi(2^{k}(x-y))2^{-2sk+2nk/p}\widehat{\psi}(2^{k}y)^{2}dy\right|^{p}dx\\ &=2^{-skp+2nk}\int_{\mathbb{R}^{n}}\left|\int_{\mathbb{R}^{n}}\varphi(2^{k}x-y)\widehat{\psi}(y)^{2}dy\right|^{p}dx\\ &=2^{-skp+nk}\int_{\mathbb{R}^{n}}\left|\int_{\mathbb{R}^{n}}\varphi(x-y)\widehat{\psi}(y)^{2}dy\right|^{p}dx \end{align*} So $\|f_{k}^{2}\|_{W^{s,p}}\gtrsim 2^{k(\frac{n}{p}-s)}\rightarrow\infty$, as $k\rightarrow\infty$. Maybe there's a soft analysis argument from which one can obtain the existence of the desired function $f$, but since the square operator is nonlinear, it's not clear to me what this argument would be.