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Suppose the claimed Poincare inequality with a partial Dirichlet boundary fails. There are $v_k\in V$ withAfter the normalization $w_k=v_k/\Vert v_k{\Vert }_{L^2(U)}$,Thus $(w_k)$ is bounded in the Sobolev space $H^1(U)$. The Rellich-Kondrashov compactness theorem and the corresponding compact embedding for a bounded $C^1$ domain give a subsequence that converges strongly in $L^2(U)$ and weakly in $H^1(U)$ to some $w$. The Sobolev space with a partial Dirichlet condition $V$ is a closed vector subspace, hence weakly closed, so $w\in V$. Moreover $\nabla w=0$, and connectedness of $U$ makes $w$ a constant function. Its trace vanishes on the positive-measure set ${\Gamma }_1$, so that constant is zero. This contradictsTherefore some $C_P$ satisfiesSince the reverse bound $\Vert \nabla v{\Vert }_2\le \Vert v{\Vert }_{H^1}$ is immediate,The gradient seminorm is a norm on $V$ because equality to zero would make $v$ a constant whose trace on ${\Gamma }_1$ is zero.

On the Hilbert space $V$, the form $B$ from part b obeysso it is a bounded bilinear form and a coercive bilinear form. The Cauchy-Schwarz inequality and the Poincare inequality with a partial Dirichlet boundary giveso $\ell$ is a bounded linear functional. The Lax-Milgram theorem now gives a unique weak solution $u\in V$. Taking $v=u$ in the weak identity yieldsand therefore

Let $U$ be a bounded Lipschitz domain and ${\Gamma }_D$ a measurable portion of its boundary. The spaceis a closed vector subspace of the H1 space. The Sobolev trace operator maps $H^1(U)$ continuously into $L^2(\partial U)$, and restriction to ${\Gamma }_D$ is bounded, so $V$ is the kernel of a continuous linear map. Boundary regularity is what makes this ordinary trace formulation available.

No positive-measure assumption is needed for closedness. If $U$ is also connected and ${\Gamma }_D$ has positive surface measure, the Poincare inequality with a partial Dirichlet boundary controls the full Sobolev norm by the gradient norm on $V$.

Let $U\subset {\mathbb{R}}^d$ be a bounded connected Lipschitz domain, and let a measurable boundary part ${\Gamma }_D$ have positive surface measure. Impose a homogeneous Dirichlet boundary condition there and a homogeneous Neumann boundary condition on the remainder. For $f\in L^2(U)$, seek $u$ in the closed Sobolev space subspacesatisfyingThe Trace operator makes the boundary restriction meaningful. The Poincare inequality with a partial Dirichlet boundary bounds the full $H^1$ norm by the gradient norm on $V$. Thus the left side is a bounded coercive bilinear form, and the Lax-Milgram theorem gives a unique weak solution. The same proof allows any bounded linear functional on $V$ as the right side.

The Neumann condition is natural: integration by parts leaves the boundary term on the non-Dirichlet portion, and requiring the identity for all $v\in V$ makes that term vanish. If ${\Gamma }_D$ is empty, constants remain in the kernel of a linear map; for homogeneous Neumann data, compatibility requires ${\int }_{U}f=0$, and uniqueness requires a normalization such as zero mean. Positive Dirichlet surface measure is what removes this ambiguity.