# 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 $$P$$.

• 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). May 24 at 11:51
• The eigenfunction is normalized, i.e. $\| u\|_{L^2(P)}=1$ May 24 at 11:53
• 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? May 24 at 11:56

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 $$\begin{eqnarray*} \|\partial_\nu u\|_{L^\infty(\partial P)} & = & |a'(1)|=\left|\int_0^1(ra'(r)'dr\right| \\ & = & \lambda\left|\int_0^1ra(r)dr\right|\le\frac\lambda{2\sqrt\pi}\,\|u\|_{L^2(P)}. \end{eqnarray*}$$
• 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 $N$. May 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 $$|P|=N|P_0|$$, 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.