Poincaré inequality for a subspace of $H^1(\Omega)$ The following is the well known Poincaré inequality for $H_0^1(\Omega)$:  

Suppose that $\Omega$ is an open set in $\mathbb{R}^n$ that is bounded. Then there is a constant $C$ such that
  $$
\int_\Omega u^2\ dx\leq C\int_\Omega|Du|^2\ dx\quad \textrm{ for all   }\ \ \color{red}{u\in H_0^1(\Omega)}.
$$

Here is my questions:  
Can the condition $u\in H_0^1(\Omega)$ be relaxed so that the inequality is still true? More precisely, suppose $H$ is a subspace of $H^1(\Omega)$ such that for any $u\in H$, $u|_\Gamma=0$, where $\Gamma$ is a ([Added:]nonempty) closed proper subset of the boundary $\partial\Omega$ and the idenity is in the sense of trace. In other words, $u$ is only zero on a part of the boundary. Do we still have the Poincaré ineqaulity in $H$?

[Added:] This question is motivated by the following exercise




 A: I'm going to augment your statement with three assumptions in order to make the proof work.  First, I'm going to assume that $\partial \Omega$ is Lipschitz.  We're going to need that to invoke the usual trace theory in $H^1(\Omega)$.  Second, I'm going to assume that $\Gamma \subseteq \partial \Omega$ is such that $\mathcal{H}^{n-1}(\Gamma) >0$, i.e. $\Gamma$ has nonzero Hausdorff measure on the boundary.  Third, I'm going to assume that $\Omega$ is connected.
Now we modify one of the standard proofs of the Poincaré inequality, namely the proof by contradiction.  Suppose, by way of contradiction, that the inequality fails.  Then for each $n \ge 1$ we can find $u_n \in H^1$ such that $u_n =0$ on $\Gamma$ and
$$
\Vert u_n \Vert_{L^2} \ge n \Vert D u_n \Vert_{L^2}.
$$
Exploiting the homogeneity of the norm, we can assume WLOG that $\Vert u_n \Vert_{L^2} = 1$ for all $n$. Then
$$
\Vert D u_n \Vert_{L^2} \le \frac{1}{n} \text{ and } \Vert u_n \Vert_{L^2} =1,
$$
from which we deduce that $\{u_n\}_{n=1}^\infty$ is uniformly bounded in $H^1(\Omega)$.  Using weak compactness and Rellich's theorem we can extract a subsequence $u_{n_k} \rightharpoonup u$ weakly in $H^1$ such that $u_{n_k} \to u$ strongly in $L^2$.
It's clear that the limit satisfies
$$
\Vert D u \Vert_{L^2} =0, \Vert u \Vert_{L^2} =1, \text{ and } u = 0 \text{ on } \Gamma.
$$
The last equality can be verified using convexity and its relation to weak convergence, for instance.  The first and third conditions, together with the fact that $\Gamma$ has positive measure and $\Omega$ is connected, tell us that $u=0$, which contradicts the fact that $\Vert u \Vert_{L^2}=1$.
Thus there exists $C >0$ such that
$$
\Vert u \Vert_{L^2} \le C \Vert D u \Vert_{L^2} \text{ for all } u \in H^1 \text{ s.t. } u\vert_{\Gamma}=0.
$$
Note that we really need connectedness here, as otherwise we can take two disconnected open sets and look at $\Gamma$ only in the boundary of one of them.  The inequality clearly fails in this case because it fails in the set that $\Gamma$ doesn't touch.
EDIT: Requested extra details.
Trace theory shows that there exists $C>0$ such that
$$
\Vert u\Vert_{L^2(\partial \Omega)} \le C \Vert u \Vert_{H^1(\Omega)}
$$
for all $u \in H^1(\Omega)$.  Note that this uses the fact that the boundary is Lipschitz.  In particular we can further bound
$$
\Vert u\Vert_{L^2(\Gamma)} \le C \Vert u \Vert_{H^1(\Omega)}.
$$
This bound shows that the set $\{u \in H^1(\Omega) : u=0 \text{ on }\Gamma\}$ is strongly closed in $H^1(\Omega)$.  However, it's also clearly a convex set.  We know from functional analysis (ultimately due to Hahn-Banach) that strongly closed convex sets are also weakly closed, so the fact that $\{u_{n_k}\}_k$ converges weakly to $u$ shows that $u$ belongs to this same set, i.e. $u=0$ on $\Gamma$.  Note, though, that this is not the only way to prove this result.
