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}.