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The following startles me. Let $\Omega \subseteq \mathbb R^n$ and write $W^{\sigma,1}(\Omega)$ for the fractional Sobolev space with norm

$$|u|_{W^{\sigma,1}(\Omega)} := \iint \frac{|u(x) - u(y)|}{|x-y|^{n+\sigma}} dx dy$$

Now, suppose that $u \in W^{\sigma,1}(\Omega)$ and that $v \in C^{0,\sigma}(\Omega)$ is bounded and Hoelder-continuous with parameter $\sigma \in (0,1)$. We may naturally ask:

When does $u v \in W^{\sigma,1}(\Omega)$ hold for the pointwise product?

If we assume that $\Omega \subseteq \mathbb R^n$ is bounded for the sake of simplicity, then a reasonable attempt at proving this inclusion looks like: $$ \iint \frac{|u(x)v(x) - u(y)v(y)|}{|x-y|^{n+\sigma}} dx dy \\ \leq \iint \frac{|u(x)v(x) - u(x)v(y)|}{|x-y|^{n+\sigma}} dx dy + \iint \frac{|u(x)v(y) - u(y)v(y)|}{|x-y|^{n+\sigma}} dx dy \\ \leq \int |u(x)| \int \frac{|v(x) - v(y)|}{|x-y|^{n+\sigma}} dy dx + \int |v(y)| \int \frac{|u(x) - u(y)|}{|x-y|^{n+\sigma}} dx dy \\ \leq \int |u(x)| \int \frac{|v(x) - v(y)|}{|x-y|^{n+\sigma}} dy dx + |v|_{\max} \iint \frac{|u(x) - u(y)|}{|x-y|^{n+\sigma}} dx dy \\ \leq |v|_{C^{\sigma}(\Omega)} \int |u(x)| \int \frac{|x - y|^\sigma}{|x-y|^{n+\sigma}} dy dx + |v|_{\max} \iint \frac{|u(x) - u(y)|}{|x-y|^{n+\sigma}} dx dy . $$ The singularity of $|x-y|^{-n}$ is too strong for convergence.

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  • $\begingroup$ You are right, I confused this. $\endgroup$
    – shuhalo
    Commented Apr 26, 2023 at 13:21

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It is not true that $W^{\sigma,1} \times C^{0,\sigma} \hookrightarrow W^{\sigma,1}$ with $\Omega = \mathbb{R}^n$.

My reference for such questions is

Thomas Runst, and Winfried Sickel. Sobolev spaces of fractional order, Nemytskij operators, and nonlinear partial differential equations. de Gruyter, 1996.

For $\sigma \in (0,1)$ and $\Omega = \mathbb{R}^n$, I understand from page 14 (and previous ones) that your $C^{0,\sigma}$ corresponds to the Besov space $B^s_{\infty,\infty}$, and that your $W^{\sigma,1}$ corresponds to the Besov space $B^s_{1,1}$.

So now the question is whether you can perform pointwise multiplication between these two Besov spaces. This is treated in Chapter 4 which gives many sufficient conditions. Section 4.3.1 gives necessary conditions for such a multiplication result to hold. In particular, you situation satisfies all the necessary conditions of page 152, except the one named (11).

The same condition prevents a similar result for $W^{\sigma,p}$ with $p \in (1,\infty)$.

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  • $\begingroup$ Thanks. As for your question, the claim 'it works for p > 1' is still within the scope of the simplifying assumption that the domain is bounded. $\endgroup$
    – shuhalo
    Commented Apr 26, 2023 at 12:15
  • $\begingroup$ I am also surprised, I would expect that one needs Hölder regularity $\tau > \sigma$. Possibly that is connected to condition (11), cannot check right now. $\endgroup$
    – Hannes
    Commented Apr 26, 2023 at 12:18
  • $\begingroup$ Condition (11) states that when you want $B^{s_1}_{p_1,q_1} \times B^{s_2}_{p_2,q_2} \hookrightarrow B^s_{p,q}$ with $s = s_1 = s_2$, you need $q \geq q_1$ and $q \geq q_2$. $\endgroup$
    – cs89
    Commented Apr 26, 2023 at 12:22
  • $\begingroup$ @Hannes: you were right, I confused some calculation. The condition $\tau > \sigma$ seems necessary. $\endgroup$
    – shuhalo
    Commented Apr 26, 2023 at 13:23
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    $\begingroup$ @shuhalo So the situation seems settled: as claimed in the book, the pointwise product does not work for the parameters of your question. $\endgroup$
    – cs89
    Commented Apr 26, 2023 at 13:52

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