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.