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.



Is the product of $u \in W^{\sigma,1}(\Omega)$ and $v \in C^{0,\sigma}(\Omega)$ again in $W^{\sigma,1}(\Omega)$?
**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 $p \in (1,\infty)$.


