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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.

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  • $\begingroup$ Suppose that $M=\partial B$ where $B$ is the unit ball. Then you want to compute $\int_{\partial B} \frac{y \cdot n}{|x-y|^{n-2}} d\sigma{y}$ $\endgroup$
    – username
    Commented Dec 29, 2013 at 15:50
  • $\begingroup$ 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. $\endgroup$
    – guacho
    Commented Jan 19, 2014 at 18:08

1 Answer 1

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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 $H^1$.

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  • $\begingroup$ Thanks, what is this bound called, because I want to look at its proof. $\endgroup$
    – soup
    Commented Dec 28, 2013 at 13:17
  • $\begingroup$ Mmm, in the case $M=\mathbb{T}$ (or $M=\mathbb{R}$), the bound can be obtained using Kato-Ponce and the Sobolev embedding. $\endgroup$
    – guacho
    Commented Dec 28, 2013 at 19:32

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