# An inequality for eigenvalues of the Dirichlet problem

Is either of these inequalities true? $$\lambda(tA + (1-t)B)\geq t\lambda(A) + (1-t)\lambda(B)$$ or $$\lambda(tA + (1-t)B)\leq t\lambda(A) + (1-t)\lambda(B),$$ where $0\leq t \leq 1$, $A,B$ are bounded domains in $\mathbb{R}^n$ and $\lambda(\Omega)$ is an eigenvalue of the problem $$-\Delta\,u=\lambda\,u\,\,\mbox{in}\,\,\, \Omega, \, u=0\,\,\,\mbox{on}\,\,\, \partial\Omega.$$

The first inequality is clearly false because $\lambda(\Omega)$ becomes large when we make the region small. As for the second, I want to rename $A'=tA$, $B'=(1-t)B$ (and then drop the primes again). Then the claim becomes $$\lambda(A+B) \le t^3\lambda (A) + (1-t)^3 \lambda(B) ,$$ for all $0\le t\le 1$. Minimize over $t$ to rewrite this as $$\lambda(A+B) \le \frac{\lambda(A)\lambda(B)}{(\sqrt{\lambda (A)} + \sqrt{\lambda (B)})^2} . \quad\quad\quad\quad (1)$$ This is true in one dimension, with equality; recall that $\lambda([0,L])=\pi^2/L^2$ to see this. I don't know what happens in higher dimensions; preliminary attempts at easy counterexamples (rectangles ...) were unsuccessful, and if I had to guess, I would now say that (1) is probably true. The inequality does hold in the special case $A=B$ (because then $A+A\supseteq 2A$).
• Please, you could explain to me again how it obtained the expression $\lambda(A+B)\leq t^3\lambda(A) + (1-t)^3\lambda(B)$. – de Araujo Jun 4 '15 at 22:29
• @deAraujo: Use that $\lambda (tC)=t^{-2}\lambda (C)$, which just follows from the definitions: if $u(x)$ solves $-\Delta u=\lambda u$ on $C$, then $u(x/t)$ works on $tC$ and vice versa. – Christian Remling Jun 4 '15 at 22:31
• Right, note that (1) is also true, with equality, for $\lambda_k([0,L])=\pi^2k^2/L^2$, for each $k\geq 1$ integer. – de Araujo Jun 5 '15 at 0:23