A consequence of the Min-Max Principle for self-adjoint operators

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, $$1966$$.

• 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 $\lambda_1(T_1)\le \langle v, T_1 v\rangle < \lambda_1(T_2)$. Sep 18, 2021 at 0:04
• @ChristianRemling Great comment! Can you put your comment as an answer? If possible with more details in order to help the community. Sep 18, 2021 at 0:19
• @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 $\langle v, T_1 v\rangle < \langle v, T_2 v\rangle =\lambda_1 \|v\|^2$. Sep 18, 2021 at 0:32
• Min-max principle indeed yields this. What is the question then? Sep 18, 2021 at 0:40
• @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. Sep 18, 2021 at 11:38

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, $$\lambda_n(T_1)\le t<\lambda_n(T_2)$$.

• Why the second equality in $(\star)$ occurs? Thanks to the ortonality of $u_j$? Sep 18, 2021 at 17:44
• Yes, because $\lambda_n$ is the minimum ev of $T_1$ restricted to $\{ u_1, \ldots , u_{n-1} \}^{\perp}$. Sep 18, 2021 at 17:59
• 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? Sep 18, 2021 at 19:02
• @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 ...") Sep 18, 2021 at 19:08