Multiplicative trace inequality on a bounded smooth domain ($\Vert Tu{\Vert }_{L^2(\partial U)}^2\le C\Vert u{\Vert }_{L^2(U)}\Vert u{\Vert }_{H^1(U)}$)

On a bounded smooth domain, the Sobolev trace theorem can be strengthened toFor 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 ${\int }_{\partial U}{u}^2={\int }_U(\mathrm{div}X)u^2+2uX\cdot Du$, and the Cauchy-Schwarz inequality gives the estimate. Density extends it to $H^1(U)$. A bound containing only the product $\Vert u{\Vert }_2\Vert Du{\Vert }_2$ is false for nonzero constant functions. Equivalently, for every $\epsilon >0$, the Young inequality gives $\Vert Tu{\Vert }_2^2\le \epsilon \Vert Du{\Vert }_2^2+C_{\epsilon }\Vert u{\Vert }_2^2$.