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If $U$ is bounded and connected and a boundary portion ${\Gamma }_D$ has positive surface measure, thenfor every $v\in H^1(U)$ whose trace vanishes on ${\Gamma }_D$. Otherwise a normalized counterexample sequence, the Rellich-Kondrashov compactness theorem, and continuity of the trace would produce a nonzero constant with zero trace on ${\Gamma }_D$.

For a boundary portion ${\Gamma }_D\subseteq \partial U$, the spaceis the kernel of the restricted trace map, hence a closed vector subspace of $H^1(U)$.

On a bounded Lipschitz domain, the Sobolev trace operator is the continuous linear extension to $H^1(\Omega )$ of the boundary restriction of smooth functions. Continuity and density make this extension unique; its existence is the Sobolev trace theorem. The kernel prescribes a homogeneous Dirichlet boundary condition. Vanishing Sobolev trace on only part of the boundary instead gives the form domain for a mixed boundary condition, with the natural Neumann boundary condition on the remainder. On an assembly of congruent tiles, taking a Sobolev trace commutes with a constant matrix acting on the tile restrictions. This lets a transplantation theorem be proved on quadratic form domains even when corner eigenfunctions have singular classical derivatives.

For $Lu=-{\partial }_j(a^{ij}{\partial }_{i}u)+b^i{\partial }_{i}u+cu$ with smooth coefficients on a bounded smooth domain, the weak formulation of $(L+\lambda )u=f$ and conormal Robin boundary condition $a^{ij}{\partial }_{i}u{\nu }_j+\beta u=0$ isThe bilinear form is bounded by the Sobolev trace theorem. If $a^{ij}$ satisfy the positive uniformly elliptic operator inequality, the multiplicative trace inequality on a bounded smooth domain and the Young inequality show that the form is coercive when $\lambda$ is sufficiently large, even if $\beta$ and $c$ have negative parts. The Lax-Milgram theorem then gives a unique weak solution for every $f\in L^2(U)$.