# If $f \in H^{\frac 12}$ and $\varphi$ is Lipschitz, is $f\varphi \in H^{\frac 12}$ (on a Lipschitz hypersurface)?

Let $M$ be a bounded hypersurface. Let $f \in H^{\frac 12}(M)$ and let $\varphi\colon M \to \mathbb{R}$ be a Lipschitz function.

When $M=\Omega \subset \mathbb{R}^n$ an open domain, we know that the multiplication $f\varphi \in H^{\frac 12}(\Omega)$. To see this, we can show the seminorm $|f\varphi|_{H^{\frac 12}}$ is bounded by adding and subtracting the same term, using triangle inequality and switching to polar coordinates.

Now let $M$ be the boundary of a bounded Lipschitz domain $\Omega \subset \mathbb{R}^{n}$, so $M$ is a compact bounded $(n-1)$-dimensional hypersurface. How do I show that $f\varphi \in H^{\frac 12}(M)$ too? I have trouble with the following term when following the same strategy as above: $$\int_M\int_M \frac{|f(x)|^2|\varphi(x)-\varphi(y)|^2}{|x-y|^{n}}d\sigma\leq \int_M|f(x)|^2\int_M |x-y|^{2-n}d\sigma$$ and I have no idea how to bound the term $\int_M |x-y|^{2-n}d\sigma$. Remember that $d\sigma$ is the surface measure.

What assumptions do I need to get this to work? Thanks.

• Suppose that $M=\partial B$ where $B$ is the unit ball. Then you want to compute $\int_{\partial B} \frac{y \cdot n}{|x-y|^{n-2}} d\sigma{y}$ – username Dec 29 '13 at 15:50
• You can get $$\|fg\|_{H^{1/2}}\leq \|f\|_{C^1}\|g\|_{H^1}$$ by using interpolation of operators. You can see the paper by Gou & Tice, Analysis and pde, vol 6, n.2m 2013. – guacho Jan 19 '14 at 18:08

## 1 Answer

We have the bound $$|fg|_{H^{0.5}}\leq C |f|_{H^{0.5+\delta}}|g|_{H^{0.5}}.$$ So, if you have a Lipschitz function, due to the compactness of the domain, the function is $H^1$.

• Thanks, what is this bound called, because I want to look at its proof. – soup Dec 28 '13 at 13:17
• Mmm, in the case $M=\mathbb{T}$ (or $M=\mathbb{R}$), the bound can be obtained using Kato-Ponce and the Sobolev embedding. – guacho Dec 28 '13 at 19:32