1
$\begingroup$

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, $1966$.

$\endgroup$

0

You must log in to answer this question.