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

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 $L^{\infty}(\mathbb{R}^{n})$.