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](https://math.stackexchange.com/questions/4246725/a-consequence-of-the-min-max-principle-for-self-adjoint-operators), 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, $1966$.



A consequence of the Min-Max Principle for self-adjoint operators
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, $\lambda\_n(T\_1)\le t<\lambda\_n(T\_2)$.


