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 $$-\Delta\,u=\lambda\,u\,\,\mbox{in}\,\,\, \Omega, \, u=0\,\,\,\mbox{on}\,\,\, \partial\Omega.$$
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2 Answers
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Update: I intended this to be a complete answer originally, but my "counterexample" was based on a miscalculation. So this is now more a collection of remarks on what I think the question is about.
We need to be more specific about which eigenvalue we want to take. I will discuss the ground state energy, which seems natural.
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 $A+A\supseteq 2A$).
answered Jun 3, 2015 at 0:22
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$\begingroup$ Please, Christian Remling, you can give me some reference. $\endgroup$ Commented Jun 4, 2015 at 22:17
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$\begingroup$ @deAraujo: References on what exactly? I'm not really using anything in this answer. $\endgroup$ Commented Jun 4, 2015 at 22:22
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$\begingroup$ Please, you could explain to me again how it obtained the expression $\lambda(A+B)\leq t^3\lambda(A) + (1-t)^3\lambda(B)$. $\endgroup$ Commented Jun 4, 2015 at 22:29
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$\begingroup$ @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. $\endgroup$ Commented Jun 4, 2015 at 22:31
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$\begingroup$ 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. $\endgroup$ Commented Jun 5, 2015 at 0:23
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The inequalities you state cannot hold. It is known that $\lambda(tA) = t^{-2}\lambda(A)$ so choosing $A = sB$ you would get $$ \frac{1}{(ts+1-t)^2}\lambda(B) \leq \geq (t/s^2+1-t)\lambda(B)$$ Neither of the inequalities can hold for any $s$. Take $s=0.5$ and $s\to 0$.
However, a Brunn-Minkowski inequality holds for the $1$-homogeneous function $\lambda^{-1/2}$. In the paper https://core.ac.uk/download/pdf/81213094.pdf and the references therein you can see that the following inequality holds $$ \lambda(tA+(1-t)B)^{-1/2} \geq t \lambda(A)^{-1/2}+(1-t)\lambda(B)^{-1/2} $$ Therefore, an inequality of the type that you want holds for $\lambda^{-1/2}$. Equality cases are also described in the paper above, for the case when $A$ and $B$ are convex.
answered Sep 29, 2021 at 15:38