5
$\begingroup$

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 $$-\Delta\,u=\lambda\,u\,\,\mbox{in}\,\,\, \Omega, \, u=0\,\,\,\mbox{on}\,\,\, \partial\Omega.$$

$\endgroup$

2 Answers 2

4
$\begingroup$

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 $A+A\supseteq 2A$).

$\endgroup$
5
  • $\begingroup$ Please, Christian Remling, you can give me some reference. $\endgroup$
    – de Araujo
    Jun 4, 2015 at 22:17
  • $\begingroup$ @deAraujo: References on what exactly? I'm not really using anything in this answer. $\endgroup$ Jun 4, 2015 at 22:22
  • $\begingroup$ Please, you could explain to me again how it obtained the expression $\lambda(A+B)\leq t^3\lambda(A) + (1-t)^3\lambda(B)$. $\endgroup$
    – de Araujo
    Jun 4, 2015 at 22:29
  • $\begingroup$ @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. $\endgroup$ Jun 4, 2015 at 22:31
  • $\begingroup$ 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. $\endgroup$
    – de Araujo
    Jun 5, 2015 at 0:23
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.