Poincare inequality on $H^1_0(M)$

Question

Is it possible to deduce the Poincare inequality for functions in $H^1_0(M)$ from the Poincare inequality for functions in $H^1(M)$ with mean value 0?

$M$ is a hypersurface with non-empty boundary.

Answer 1

I don't think so. The zero-mean inequality reflects the interconnectivity of $M$, how easy it is to move between different parts of it. The zero-boundary inequality measures how large and easily accessible the boundary of $M$ is. These are quite different things.

On one hand, an hourglass-shaped surface, without top and bottom lids, admits a nice zero-boundary inequality, but a lousy zero-mean inequality (I'm measuring niceness by size of constant). The latter is because a function can be $1$ in the bottom half and $-1$ in the upper half, with transition in the narrow part of the surface.

On the other hand, a sphere with a little disk cut out of it has a nice zero-mean inequality (pretty much the same as for the whole sphere), but a lousy zero-boundary inequality. Indeed, let $B(a,r)$ be the removed disk, let $R>r$ (but keep $R$ small compared to the radius of the sphere), and define $$u(x)=\begin{cases}0 \quad & x\in B(a,r)\\ 1 \quad &x\notin B(a,R) \\ \frac{\log d(x,a)-\log r}{\log R-\log r} \quad & x\in B(a,R)\setminus B(a,r)\end{cases} $$ The energy integral of $u$ is $\int|\nabla u|^2 \approx \frac{2\pi}{\log R-\log r}$. This can be made arbitrarily small by keeping $R$ fixed and making $r$ much smaller than $R$.

Thus, the constant in zero-boundary inequality for $S^2\setminus B(a,r)$ blows up as $r\to 0$. Yet, the constant in the zero-mean inequality is about the same as for $S^2$ itself (I don't have a proof but this is "obvious", isn't it?). So it does not look like one could get the zero-boundary inequality from zero-mean.