= Multiplicative trace inequality on a bounded smooth domain
{title2=$\|Tu\|_{L^2(\partial U)}^2\leq C\|u\|_{L^2(U)}\|u\|_{H^1(U)}$}
On a bounded smooth domain, the <Sobolev trace theorem> can be strengthened to
$$
\|Tu\|_{L^2(\partial U)}^2\leq C\bigl(\|u\|_{L^2(U)}^2+\|u\|_{L^2(U)}\|Du\|_{L^2(U)}\bigr)
\leq C'\|u\|_{L^2(U)}\|u\|_{H^1(U)}.
$$
For smooth $u$, extend the outward unit normal to a smooth vector field $X$ near the closure of $U$ with $X\cdot\nu=1$ on its boundary. The <divergence theorem> gives $\int_{\partial U}u^2=\int_U(\operatorname{div}X)u^2+2uX\cdot Du$, and the <Cauchy-Schwarz inequality> gives the estimate. Density extends it to $H^1(U)$. A bound containing only the product $\|u\|_2\|Du\|_2$ is false for nonzero constant functions. Equivalently, for every $\varepsilon>0$, the <Young inequality> gives $\|Tu\|_2^2\leq\varepsilon\|Du\|_2^2+C_\varepsilon\|u\|_2^2$.
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