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Let $\Omega$ be a fixed, smooth and bounded domain in $\mathbb{R}^n$. If we denote with $\{\lambda_n\}_{n \ge 1}$ the nondecreasing sequence of eigenvalues of the Dirichlet problem $$\left\{ \begin{array}{rc} -\Delta u = \lambda u & \Omega \\ u=0 & \partial\Omega \end{array}\right.$$ and we denote with $\{\mu_n\}_{n \ge 1}$ the nondecreasing sequence of eigenvalues of the Neumann problem $$\left\{\begin{array}{rc} -\Delta u = \mu u & \Omega \\ \frac{\partial u}{\partial \nu}=0 & \partial\Omega \end{array}\right.$$ then we have $$\tag{1}\mu_j\le \lambda_j,\quad \forall j=1, 2, \ldots$$ I would like to prove (1).

The page at the Encyclopaedia of Mathematics invokes the max-min characterization: both for Dirichlet and for Neumann eigenvalues we have $$\lambda_k=\sup \inf \frac{\int \lvert \nabla u\rvert^2}{\int u^2} [=\mu_k],$$ where the $\inf$ is taken over all $u\in H^1_0$ [resp. $H^1$] which are orthogonal to $\phi_1\ldots \phi_{k-1}$ and the $\sup$ is taken over all possible choices of $\phi_1\ldots \phi_{k-1}\in H^1_0$ [resp. $H^1$].

Question. According to the linked page, inequality (1) follows "obviously" from this characterization. However, this does not seem that obvious to me, as the $\sup$ for the Neumann eigenvalues is taken over a larger class. Can someone explain me how is this done?

Thank you for reading.

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  • $\begingroup$ Does anybody know about the inequalities in the opposite direction? In particular, are there any results on interlacing/alternating of the Dirichlet and Neumann eigenvalues for Laplace-Beltrami operator?... $\endgroup$
    – DVD
    Jun 12 '14 at 16:25
  • $\begingroup$ @Daved: I think that you'd better ask another question. You'd get a lot more visibility. $\endgroup$ Jun 12 '14 at 16:53
  • $\begingroup$ Thank you for the advice. I just did... $\endgroup$
    – DVD
    Jun 12 '14 at 17:58
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It has to do with the way the different boundary value problems are formulated. In one, you are considering functions lying in $H^1_0$, in the other, over functions lying in $H^1$. Recall that $H^1_0$ is a subset of $H^1$.

(edited after commentary): Aha, I realized. My answer is still correct, but the formulation needs to be revised. The sup is taken over all $\phi$ in $H^1$ in both cases, in which case it's only the conditions on the inf part that change between the Dirichlet and Neumann eigenvalues, which lets you compare apples to apples, so to speak.

Taking $$\lambda_k = \sup \inf \frac{\int |\nabla u|^2}{\int u^2}$$ where we take $u\in H^1_0$ orthogonal to $\phi_1 \ldots \phi_{k-1}$ lying inside $H^1$ still yields the traditional Dirichlet eigenvalue. It is clear that when the $\phi_i$ are the $k-1$ Dirichlet eigenfunctions, $\lambda_k$ is achieved by the ratio $$D[\phi_1,\ldots,\phi_{k-1}] = \inf \frac{\int |\nabla u|^2}{\int u^2}$$ We need only show that for any other choice of the $\phi_i$, the ratio achieved is $\leq \lambda_k$. Morally, the argument goes like this: suppose the space of functions spanned by the $\phi_i$ does not include the eigenspace corresponding to $\lambda_1$; then the natural value for the ratio is $\lambda_1 \leq \lambda_k$. You repeat inductively on each of the eigenspaces corresponding to eigenvalues, until you have specified $k-1$ functions, and then what's left must be $\lambda_k$.

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  • $\begingroup$ Yes, but this does not help us IMHO. Since $H^1_0\subset H^1$, taking the sup over $\phi_1\ldots \phi_{n-1}\subset H^1_0$ yields a smaller number that taking the sup over $H^1$. However, we are trying to prove the opposite inequality. $\endgroup$ Apr 1 '13 at 11:34
  • $\begingroup$ Aha. Let's work inductively. The argument works for the first eigenvalue, since there are no $\phi$s to be considering for that part. Let me think a bit about the rest ... $\endgroup$
    – Ray Yang
    Apr 1 '13 at 14:36
  • $\begingroup$ Aha. This works much more cleanly if you realize the sup inf definition still works when the $\phi_i$ are in $H^1$ in both cases. I've put a more detailed explanation in above. $\endgroup$
    – Ray Yang
    Apr 2 '13 at 16:22
  • $\begingroup$ Now you have convinced me! Thank you very much. $\endgroup$ Apr 3 '13 at 8:48

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