First Dirichlet eigenvalue on regular polygons Assume $\lambda(P)$ is the first Dirichlet eigenvalue of a regular polygon $P$. Let $u$ be the corresponding eigenfunction, normalized by $\|u\|_{L^2(P)}=1$,  and $\partial_{\nu}u$ be its normal derivative on the boundary. Is the following estimate correct:
\begin{eqnarray*}
\lambda(P)\geq \|\partial_{\nu} u\|^2_{L^{\infty}(\partial P)}\vert P\vert\quad?
\end{eqnarray*}
Here $\vert P\vert$ denotes the area of $P$.
 A: Ignoring the normalization, your inequality writes
$$\lambda(P)\|u\|_{L^2(P)}^2\ge\|\partial_\nu u\|_{L^\infty(\partial P)}^2|P|\qquad?$$
Good news: this is scaling invariant.
Bad news: this is false for a square $P=(-\pi/2,\pi/2)^2$. Then $u(x,y)=\cos x\cos y$,$\lambda(P)=2$, $|P|=\pi^2$ and
$$\|u\|_{L^2}^2=\left(\int_{-\pi/2}^{\pi/2}\cos^2 x\,dx\right)^2=\frac{\pi^2}4\,,\qquad\|\partial_\nu u\|_{L^\infty}^2=1.$$
Perhaps your inequality is correct with an extra constant factor.Perhaps also you should think to an inequality of the form
$$\lambda(P)\|u\|_{L^2(P)}\ge c\|\partial_\nu u\|_{L^\infty(\partial P)},$$
which is scaling invariant too. For instance the limit case of the unit disk gives $u(x,y)=a(r)$ with
$$a''+\frac1r\,a'=-\lambda a \quad a'(0)=0,\quad a(1)=0.$$
Then
\begin{eqnarray*}
\|\partial_\nu u\|_{L^\infty(\partial P)} & = & |a'(1)|=\left|\int_0^1(ra'(r)'dr\right| \\
& = & \lambda\left|\int_0^1ra(r)dr\right|\le\frac\lambda{2\sqrt\pi}\,\|u\|_{L^2(P)}.
\end{eqnarray*}
A: There is a more general reason why any such statement will fail: If you consider a $P$ consisting of $N$ separate copies of the same basic region $P_0$, then $|P|=N|P_0|$, while everything else in your inequality is independent of $N$, so the inequality will fail for large $N$. (This problem has a degenerate ground state, but of course you could address this by slightly changing the shapes.)
If you connect the components by thin tubes, then you now have a connected $P$ and are still approximately in the situation described above.
