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Let $\Omega \subseteq \mathbb{R}^d$ be bounded and open with $\partial\Omega$ Lipschitz, $1<p<\infty$.

My question: knowing that $f,g\in W^{1,p}(\Omega)$ for what $p$ can we conclude that $f\cdot g\in W^{1,p}(\Omega)$?
In other words, for what $p\in (1,\infty)$ is there a constant $c>0$ such that:

$$\Vert fg\Vert_{W^{1,p}(\Omega)}\leq c\Vert f\Vert_{W^{1,p}(\Omega)}\cdot \Vert g\Vert_{W^{1,p}(\Omega)},\quad \forall\ f,g\in W^{1,p}(\Omega)\;?$$

So, is $W^{1,p}(\Omega)$ a Banach algebra?

P.S. I know that when $d=1$ the answer is yes, and it is proved in Brezis book, page 214. I also know that in some lecture notes of Terence Tao (lecture 4 here) it is stated that this is true when $p>d$ for $\Omega=\mathbb{R}^d$.

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    $\begingroup$ The point is that if $p > d$, $$ W^{1,p}(\Omega) \subset L^\infty(\Omega)$$ and there exists $C > 0$ such that $$ \|f\|_\infty \le C\|f\|_{W^{1,p}}. $$ Therefore, \begin{align*} \|\nabla(fg)\|_{W^{1,p}} &= \|f\nabla g + g\nabla f\|_{W^{1,p}}\\ & \le \|f\|_\infty\|g\|_{W^{1,p}} + \|g\|_{\infty}\|f\|_{W^{1,p}} \\ & \le 2C\|f\|_{W^{1,p}}\|g\|_{W^{1,p}}. \end{align*} I do not believe you can do better than that and can find counterexamples. $\endgroup$
    – Deane Yang
    Commented Sep 16, 2023 at 23:04
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    $\begingroup$ A general rule of thumb (but not a formal theorem) is that a Sobolev space should be a Banach algebra iff it embeds into $L^\infty$ (because the spectrum should coincide with the essential range). $\endgroup$
    – Terry Tao
    Commented Sep 17, 2023 at 1:23
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    $\begingroup$ This is proven in the book by Adams and Fournier, "Sobolev spaces" $\endgroup$
    – Romain Gicquaud
    Commented Sep 19, 2023 at 9:35

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I'll just complete the answers in the comments showing that for $p\le d$ it is not a Banach algebra. Let me take $\Omega=B_1(0) \subset \mathbb{R}^n$ and $p=d$.

Assume that $W^{1,d}(B_1) $ is a Banach algebra and that the inequality $$\| u\cdot v \|_{W^{1,n}(B_1)} \le C_0 \| u \|_{W^{1,n}(B_1)} \| v\|_{W^{1,n}(B_1)}$$ holds for every $u,v \in W^{1,n}(B_1)$. Choosing $u=v$ and iterating the inequality gives you $ \| u^k\|_{W^{1,n}(B_1)} \le C_0^k \| u \|_{W^{1,n}(B_1)}^k$ for every $k\ge 0$. This would imply that $e^u \in W^{1,n}(B_1) $ since, by the inequality we have just obtained, you have absolute convergence of the series $\sum_{k\ge 0} \frac{u^k}{k!}$ in $W^{1,n}(B_1)$. Now, the fact $u\in W^{1,n}(B_1) \implies e^u \in W^{1,n}(B_1)$ is clearly not true and you can find tons of counterexamples (e.g. $u(x)=\log\log(1/|x|)$).

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