The Poincaré-constant of a domain ${\Omega\subset\mathbb{R}^{n}}$ is the smallest constant ${C}$ such that the estimate

$\displaystyle \|u- \bar u\|_{p}\leq C\|\nabla u\|_{p}$

holds (where ${\bar u = |\Omega|^{-1}\int_{\Omega} u}$ is the mean value of ${u}$).

These constants are known for some classes of domains and some values of ${p}$: E.g. Payne and Weinberger showed in 1960 that for ${p=2}$ and convex ${\Omega}$ the constant is ${\tfrac{\mathrm{diam}(\Omega)}{\pi}}$ and Acosta and Duran showed in 2004 that for ${p=1}$ and convex ${\Omega}$ one gets ${\tfrac{\mathrm{diam}(\Omega)}2}$.

I do not know of any other results on these constants, and while playing around with these kind of things, I derived another one: Since I often work with images ${u:\Omega\rightarrow [0,1]}$ I will assume that ${0\leq u\leq 1}$ in the following.

Using the co-area formula we can express the ${1}$-norm of the gradient via the length of the level sets: Denoting by ${\mathcal{H}^{n-1}}$ the ${n-1}$-dimensional Hausdorff measure (which is, roughly speaking, the length in the case of ${n=2}$) and with ${E_{t}}$ the level set of ${u}$ at level ${t}$, the co-area formula states that

$\displaystyle \int_{\Omega}|\nabla u| = \int_{0}^{1}\mathcal{H}^{n-1}(\partial E_{t}) dt. \ \ \ \ \ (1)$

Now we combine this with the isoperimetric inequality, which is

$\displaystyle \mathcal{H}^{n-1}(\partial E_{t})\geq n |B_{1}|^{\tfrac1n}|E_{t}|^{\tfrac{n-1}{n}}, \ \ \ \ \ (2)$

where ${B_{1}}$ is unit ball and ${|\cdot|}$ denotes the Lebesgue measure. Combining~(1) and~(2) we get

$\displaystyle \int_{\Omega}|\nabla u| \geq n |B_{1}|^{\tfrac1n}\int_{0}^{1}|E_{t}|^{\tfrac{n-1}{n}} dt. \ \ \ \ \ (3)$

Now we use the trivial fact that

$\displaystyle u(x)^{2} = \int_{0}^{u(x)}2t dt = \int_{0}^{1}2t\chi_{\{x\mid u(x)\geq t\}}(x)dt$

combined with Fubini to get

$\displaystyle \int_{\Omega}u^{2} = \int_{0}^{1}2t\int_{\Omega}\chi_{\{x\mid u(x)\geq t\}}(x)dx\, dt = 2\int_{0}^{1}t |E_{t}|dt. \ \ \ \ \ (4)$

Since ${|E_{t}|/|\Omega|\leq 1}$ and ${(n-1)/n<1}$ it holds that ${|E_{t}|\leq |\Omega|^{\tfrac1n}|E_{t}|^{\tfrac{n-1}n}}$. Combining this with~(4) we get

$\displaystyle \int_{\Omega}u^{2}\leq 2|\Omega|^{\tfrac1n}\int_{0}^{1}t |E_{t}|^{\tfrac{n-1}n}dt \leq 2|\Omega|^{\tfrac1n}\int_{0}^{1} |E_{t}|^{\tfrac{n-1}n}dt.$

Finally, we use~(3) to obtain

$\displaystyle \int_{\Omega}u^{2}\leq \frac{2|\Omega|^{\tfrac1n}}{n|B_{1}|^{\tfrac1n}}\int_{\Omega}|\nabla u|$

in other words

$\displaystyle \|u\|_{2}^{2}\leq \frac{2|\Omega|^{\tfrac1n}}{n|B_{1}|^{\tfrac1n}}\|\nabla u\|_{1}.$

This estimate is quite explicit, does not need the subtraction of the mean value, does not need convexity of ${\Omega}$, but also does not obey the scaling ${u\mapsto au}$ (which is of no surprise since we used the condition ${0\leq u\leq 1}$ which also does not obey this scaling).

In dimension ${n=2}$ the estimate takes the simpler form

$\displaystyle \|u\|_{2}^{2}\leq \sqrt{\tfrac{|\Omega|}{\pi}}\|\nabla u\|_{1}.$