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 $P$.

$\begingroup$ Since you can multiply an eigenfunction by a non zero constant and still have an eigenfunction, your estimate cannot hold (by making this constant arbitrary large). $\endgroup$– HéhéhéMay 24 at 11:51

$\begingroup$ The eigenfunction is normalized, i.e. $\ u\_{L^2(P)}=1$ $\endgroup$– guest61May 24 at 11:53

1$\begingroup$ You should edit your question then because it does not appear. By the way, I guess that you are talking about eigenvalue of the laplacian operator but you didn't write it down. Moreover, it should be $\ \partial_\nu u \$ rather than $\ \nabla u\$, shoudn't it? $\endgroup$– HéhéhéMay 24 at 11:56
2 Answers
Ignoring the normalization, your inequality writes $$\lambda(P)\u\_{L^2(P)}^2\ge\\partial_\nu u\_{L^\infty(\partial P)}^2P\qquad?$$ 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*}

$\begingroup$ Many thanks for the detailed answer and the counter example. Unfortunately I need the constant $1$ and the power $2$. For the disk the desired estimate holds with equality. Maybe there is some hope that it holds for a regular $N$ polygon for $N$ large enough. Otherwise the reverse inequality may have a chance for all $N$. $\endgroup$– guest61May 24 at 14:23
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 $P=NP_0$, 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.