# 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 $$\lambda_n(T_1) < \lambda_n(T_2)$$ for each $$n \in \{1,\cdots, N\}$$?

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

$$[1]$$ 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, $$1966$$.