Poincaré inequality on metric spaces

140 Views Asked by At

In this book, page 91, example 4.18, it is said that the space $$Y=\{z\in\mathbb{C}:\ |z|=1,\ \arg z\in [-T,T]\}$$

where $3<T<\pi$ does not support a strong Poincare inequality (see page 84 for the definition). Does anyone knows why this is true?

They said that $Y$ does admits a weak Poincare inequality with dilation constant $\lambda\to \infty$ if $T\to \pi$, however I fail to see this. Because $\Phi$ is an isometry, it seems to me that $Y$ also admits a strong Poincare inequality. Maybe I am doing some mitake here in my calculations!

1

There are 1 best solutions below

3
On BEST ANSWER

But $\Phi(x)=e^{ix}$ is not an isometry. The space $Y$ carries the restriction metric from $\mathbb C$. Because of this, a ball $B$ centered at $e^{iT}\in Y$ looks like this (colored red):

enter image description here

One can imagine a nice function $g:Y\to \mathbb R$ that has zero gradient on $B$ but takes different values on two components of $B$. This is why the strong form of the Poincaré inequality fails.

The weak form still holds, but $\lambda B$ must be large enough to cover all of $Y$, because the gradient of $g$ could be supported on a short arc somewhere far on the right. This is why a very large $\lambda$ may be needed.