# Product of Hölder and Sobolev functions

Here $C^{\kappa , \lambda} ( \overline{\Omega} ) = \left\{ h|_{\overline{\Omega}} :h \in C^{\kappa , \lambda} ( \mathbb{R}^{n} ) \text{ and } h \text{ has compact support} \right\}$ denotes $\kappa$ times Hölder continuously differentiable functions in $\overline{\Omega}$ and $W^{s,p}(\Omega)$ is the usual Sobolev space with possibly non-integer exponent $s\geq0$.

Proposition: Let $\Omega \subset \mathbb{R}^{n}$ be open, bounded and Lipschitz. Let $s \geq 0 ,1 \leq p< \infty$ and $v \in W^{s,p} ( \Omega ) ,h \in C^{\kappa , \lambda}_{c} ( \overline{\Omega} )$ where $\kappa \in \mathbb{N}_{0} , \lambda \in ( 0,1 ]$ and $\kappa + \lambda \geqslant s$ if $s \in \mathbb{Z}$ and $\kappa + \lambda >s$ otherwise. Then the product $hu \in W^{s,p} ( \Omega )$ and there exists a constant such that $$\| hu \|_{W^{s,p} ( \Omega )} \leqslant C \| u \|_{W^{s,p} ( \Omega )}.$$

Attempt at proof: The case $s \in \mathbb{N}_0$ is easy since the Sobolev norm doesn't involve the Slobodecijk seminorm and I can use that all partial derivatives of $h$ are uniformly bounded. However, for non integer $s=\lfloor s \rfloor+t, t \in (0,1)$ this seminorm does appear:

$$\| u \|_{W^{s,p} ( \Omega )} := \sum_{| \alpha | \leqslant \lfloor s \rfloor} \| D^{\alpha} u \|^{p}_{L^{p} ( \Omega )} + \sum_{| \alpha | = \lfloor s \rfloor} \underset{\Omega \times \Omega}{\int \int} \frac{| D^{\alpha} u ( x ) -D^{\alpha} u ( y ) |}{| x-y |^{n+tp}} d x d y.$$

And I don't know how to handle the double integrals. Even in the simplest situation where $\lfloor s \rfloor=0$, I don't know how to bound above the integral

$$\underset{\Omega \times \Omega}{\int \int} \frac{| h ( x ) u ( x ) -h ( y ) u ( y ) |}{| x-y |^{n+tp}} d x d y.$$

I fear this might be some trivial inequality, but I just can't see it. Any help would be greatly welcome.

• You can probably find many good ideas in this paper: arxiv.org/abs/1104.4345 Apr 1, 2014 at 9:43

Consider the case $\kappa = 0$, $1 < s < \lambda \leqslant 1$, the general one can be easily deduced from this one. Firstly, because $h$ is bounded in $\overline{\Omega}$ we have trivially: $\| hu \|_p \leqslant \| h \|_{\infty} \| u \|_p$. Secondly, for the Gagliardo seminorm $$[ hu ]_{s, p}^p := \underset{\Omega \times \Omega}{\int \int} \frac{| h (x) u (x) - h (y) u (y) |^p}{| x - y |^{n + sp}} \mathrm{d}x \mathrm{d}y,$$ we use the convexity of $x \mapsto x^p$ twice (that is, we use $| a + b |^p \leqslant 2^{p - 1} (| a |^p + | b |^p)$ and obtain \begin{eqnarray*} \hspace{1em} {| h (x) u (x) - h (y) u (y) |^p} & \leqslant & C \left( | h (x) u (x) - h (x) u (y) |^p + | h (x) u (y) - h (x) u (x) |^p \\ \hspace{2em} + | h (x) u (x) - h (y) u (x) |^p + | h (y) u (x) - h (y) u (y) |^p \right)\\ & = & C \left( | h (x) |^p | u (x) - u (y) |^p + | h (x) |^p | u (y) - u (x) |^p \\ \hspace{2em} + | h (x) - h (y) |^p | u (x) |^p + | h (y) |^p | u (x) - u (y) |^p \right) . \end{eqnarray*} We plug this into the integral but for the third summand we use first that $$| h (x) - h (y) |^p \leqslant C | x - y |^{\lambda p},$$ then: \begin{eqnarray*} [ hu ]_{s, p}^p & \leqslant & C \left( \| h \|_{\infty}^p 3 \underset{\Omega \times \Omega}{\int \int} \frac{| u (x) - u (y) |^p}{| x - y |^{n + sp}} \mathrm{d}x \mathrm{d}y + \underset{\Omega \times \Omega}{\int \int} \frac{C | x - y |^{\lambda p} | u (x) |^p}{| x - y |^{n + sp}} \mathrm{d}x \mathrm{d}y \right)\\ & \leqslant & C \left( [ u ]_{s, p}^p + \underset{\Omega \times \Omega}{\int \int} \frac{ | u (x) |^p}{| x - y |^{n + (s - \lambda) p}} \mathrm{d}x \mathrm{d}y \right) . \end{eqnarray*} Notice that the exponent in the denominator of the integrand is $n - \delta < n$ for some $\delta > 0$ so the function $1 / | z |^{n - \delta}$ is integrable in any bounded domain. We use Fubini-Tonelli and change the variable in the $\mathrm{d}y$ integral with $z = x - y$ to obtain: \begin{eqnarray*} [ hu ]_{s, p}^p & \leqslant & C \left( [ u ]_{s, p}^p + \int_{\Omega} | u (x) |^p \int_{x - \Omega} \frac{1}{| z |^{n - \delta}} \mathrm{d}z \mathrm{d}x \right)\\ & \leqslant & C \left( [ u ]_{s, p}^p + \int_{\Omega} | u (x) |^p \int_{\Omega + \Omega} \frac{1}{| z |^{n - \delta}} \mathrm{d}z \mathrm{d} x \right)\\ & \leqslant & C ([ u ]_{s, p}^p + \| u \|_p^p) . \end{eqnarray*} Therefore $$\| hu \|_{s, p} \leqslant C \| u \|_{s, p},$$ and keeping track of the constants above we see that indeed $C = C (h, p, \Omega)$.