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Let $H=(H, (\cdot, \cdot))$ be a Hilbert space. Let $T_1,T_2:D \subset H \longrightarrow H$ be a self-adjoint operators (not necessarily bounded). It's well-know that the spectrum $\sigma(T_i)$ of $T_i$ satisfies $\sigma(T_i) \subset \mathbb{R}$, for $i=1,2$ (see Theorem $29.2$ in $[3]$). Suppose that $T_1$ and $T_2$ are bounded below and has $N \in \mathbb{N}$ (real) eigenvalues arranged in the ascending order $$ \lambda_1(T_i) \leq \lambda_2(T_i) \leq \lambda_3(T_i) \leq \cdots \lambda_N(T_i), \quad i \in \{1,2\}. $$

As a consequence of the Min-Max Principle $($see $[2$, page $85]$ or $[1$, page $61])$, if $$ (T_1(u), u) \leq (T_2(u), u),\; \forall \; u \in D \tag{1} $$ then, for each $n \in \{1,\cdots, N\}$, $$\lambda_n(T_1) \leq \lambda_n(T_2). \tag{2}$$

Question. If $$ (T_1(u), u) < (T_2(u), u),\; \forall \; u \in D\setminus \{0\} $$ and then $$\lambda_n(T_1) < \lambda_n(T_2) \tag{3} $$ for each $n \in \{1,\cdots, N\}$?

I think so, because the Min-Max Principle establishes that, for $i=1,2$, $$ \lambda_n(T_i)= \sup_{u_1, u_2, \cdots u_{n-1} \in H } \inf_{v \in D\setminus \{0\} \atop v \in [u_1, u_2, \cdots u_{n-1}]^{\perp} } \frac{(T_i(v),v)}{\|v\|}. $$

Remark. I did this question in Math Stackexchange, but I don't received any comment or answer.

Any comment or reference are welcome.

$[1]$ Kato, T., Perturbation Theory for Linear Operators, $2$nd edition, Springer, Berlin, $1984$.

$[2]$ Reed, S. and Simon, B., Methods of Modern Mathematical Physics: Analysis of Operator, Academic Press, Vol. IV, $1978$.

$[3]$ Bachman, G. and Narici, L. Functional Analysis. New York: Academic Press, $1966$.

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  • $\begingroup$ Yes, and the proof is the same as in the other case (eq. (1)): For example, if $T_2 v=\lambda_1(T_2)v$, then $\lambda_1(T_1)\le \langle v, T_1 v\rangle < \lambda_1(T_2)$. $\endgroup$ Sep 18, 2021 at 0:04
  • $\begingroup$ @ChristianRemling Great comment! Can you put your comment as an answer? If possible with more details in order to help the community. $\endgroup$
    – Guilherme
    Sep 18, 2021 at 0:19
  • $\begingroup$ @ChristianRemling But why $\lambda_1(T_1)\le \langle v, T_1 v\rangle < \lambda_1(T_2)$? The first inequality I don't see because we have $\sup \inf$ and the second because $ \langle v, T_1 v\rangle < \langle v, T_2 v\rangle =\lambda_1 \|v\|^2$. $\endgroup$
    – Guilherme
    Sep 18, 2021 at 0:32
  • $\begingroup$ Min-max principle indeed yields this. What is the question then? $\endgroup$ Sep 18, 2021 at 0:40
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    $\begingroup$ @FedorPetrov But my question is if $\lambda_n(T_1) < \lambda_n(T_2)$ holds, that is, if the strict inequality occurs. See the Question, please. $\endgroup$
    – Guilherme
    Sep 18, 2021 at 11:38

1 Answer 1

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I'm expanding my comment, in response to the OP's comment. Indeed, the case of just the lowest eigenvalue is perhaps not a good illustration of the full argument.

In general, let $u_j$ be a normalized eigenvector for $\lambda_j(T_1)$, so $T_1 u_j=\lambda_j(T_1) u_j$. Also make sure that the $u_j$ are orthogonal (this is automatic, except in the case of degeneracies). Then $$ \lambda_n(T_1)=\langle u_n, T_1 u_n\rangle =\inf_{v\perp u_1,\ldots , u_{n-1}} \langle v, T_1 v\rangle < \inf _{v\perp u_1,\ldots , u_{n-1}} \langle v, T_2 v\rangle . $$ The inequality is true because the infima are really minima: by the assumption on the existence of discrete spectrum in the range we're investigating and by the spectral theorem, the search for $v$ can be restricted to a suitable finite-dimensional subspace. We can then try the $v$ that minimizes $\langle v, T_2 v\rangle$ in the other quadratic form.

We're done since obviously $\lambda_n(T_2)\ge \inf _{v\perp u_1,\ldots , u_{n-1}} \langle v, T_2 v\rangle$.


The argument can also be organized differently, maybe this version is more transparent: Let $M\subseteq H$, $\dim M=n$, be the ("an", in the case of degeneracies) space spanned by the eigenvectors of $T_2$ with eigenvalues $\lambda_1(T_2),\ldots , \lambda_n(T_2)$. Then $\max_{v\in M, \|v\|=1 }\langle v, T_2 v\rangle = \lambda_n(T_2)$.

By assumption and since $M$ is finite-dimensional, $$ t=\max_{v\in M, \|v\|=1 } \langle v, T_1 v\rangle < \lambda_n(T_2) . $$ For any choice of $u_1,\ldots , u_{n-1}\in H$, there will be a $v\in M\ominus H$, $\|v\|=1$. Hence, by min-max, $\lambda_n(T_1)\le t<\lambda_n(T_2)$.

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  • $\begingroup$ Why the second equality in $(\star)$ occurs? Thanks to the ortonality of $u_j$? $\endgroup$
    – Guilherme
    Sep 18, 2021 at 17:44
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    $\begingroup$ Yes, because $\lambda_n$ is the minimum ev of $T_1$ restricted to $\{ u_1, \ldots , u_{n-1} \}^{\perp}$. $\endgroup$ Sep 18, 2021 at 17:59
  • $\begingroup$ And what occurs if, for instance, $\lambda_{n-1}= \lambda_{n}$? Can we to guarantee the orthogonality between $u_{n-1}$ and $u_{n}$? I think not. In this case, is the above argument still valid? $\endgroup$
    – Guilherme
    Sep 18, 2021 at 19:02
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    $\begingroup$ @Guilherme: In this case, I choose the eigenvectors as orthogonal vectors. (This is exactly what I'm addressing in the sentence "Also, make sure that ...") $\endgroup$ Sep 18, 2021 at 19:08

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