Question about the proof of the min-max principle

Question

Let $A$ be a densely defined self-adjoint operator on a Hilbert space $\mathscr{H}$ which is bounded from below. The min-max principle states that: $$\mu_{n}(A) := \inf_{\substack{M \subset D(A) \\ \text{dim}M = n}}\sup_{\substack{\psi \in M \\ \|\psi\| = 1}}\langle \psi, A\psi\rangle$$ is an increasing sequence $\mu_{n}(A) \to \mu_{\infty}(A) := \inf \sigma_{\text{ess}}(A)$. Moreover, if $\mu_{n}(A) < \mu_{\infty}(A)$ then $\mu_{n}(A)$ is the $n$-th lowest eigenvalue of $A$.

I am trying to understand the proof of the second part of this result. I already proved that $\mu_{1}(A)$ is the lowest eigenvalue of $A$ if $\mu_{1}(A) < \mu_{\infty}(A)$. Suppose then $\mu_{2}(A) < \mu_{\infty}(A)$. Since $\mu_{1}(A)$ is an eigenvalue of $A$, we can choose an associated eigenvector $\varphi$. Of course we can decompose $\mathscr{H} = \text{span}\{\varphi\}\oplus W$, where $W = \{\varphi\}^{\perp}$. Let $A_{W}$ be the restriction of $A$ to $W$. Here is my question: the proof states that: $$\mu_{2}(A) = \inf_{\substack{\psi \in W \\ \|\psi\| =1 }}\langle \psi, A_{W}\psi\rangle \equiv \mu_{1}(A_{W}).$$ Why is this true?

Answer 1

Please point exact p. or equality number to clarify.

1. $A_W:D(A_W)\to W$ is self-adjoint operator bounded from below.
2. $D(A_W)\oplus \text{span}\{\phi\} = D(A), A = A_W\oplus \mu_1(X)I$ where $I$ is identity map on $\text{span}\{\phi\}=\mathbb{C}\phi.$
3. $\mu_\infty(A)=\mu_\infty(A_W).$

\begin{align} E &:= \inf_{ \substack{M \subset D(A) \\ \text{dim}M = 2}} \sup_{\substack{\psi \in M \\ \|\psi\| = 1}} \langle \psi, A\psi\rangle \\ & \overset{1}{=} \inf_{ \substack{M \subset D(A) \\ \text{dim}M = 2}} \sup_{\substack{\psi \in M \\ \|\psi\| = 1 \\ \psi\perp\phi}} \langle \psi, A\psi\rangle \\ & \overset{2}{=} \inf_{ \substack{M \subset D(A) \\ \text{dim}M = 2}} \sup_{\substack{\psi \in M\cap W \\ \|\psi\| = 1 }} \langle \psi, A\psi\rangle \\ & \overset{3}{=} \inf_{ \substack{M \subset D(A) \cap W \\ \text{dim}M \in \{1,2\} }} \sup_{\substack{\psi \in M \\ \|\psi\| = 1 }} \langle \psi, A\psi\rangle \end{align}

Simplify last expression: \begin{align} \inf_{ \substack{M \subset D(A) \cap W \\ \text{dim}M = 2}} \sup_{\substack{\psi \in M \\ \|\psi\| = 1 }} \langle \psi, A\psi\rangle \overset{4}{\ge} \inf_{ \substack{M \subset D(A) \cap W \\ \text{dim}M = 1}} \sup_{\substack{\psi \in M \\ \|\psi\| = 1 }} \langle \psi, A\psi\rangle \end{align}

Continue: \begin{align} E &\overset{6}{=} \inf_{ \substack{M \subset D(A) \cap W \\ \text{dim}M = 1}} \sup_{\substack{\psi \in M \\ \|\psi\| = 1 }} \langle \psi, A\psi\rangle \\ &\overset{7}{=} \inf_{ \substack{M \subset D(A) \cap W \\ \text{dim}M = 1 \\ \|\psi\| = 1}} \langle \psi, A\psi\rangle \\ &\overset{8}{=} \inf_{ \substack{M \subset D(A_W) \\ \text{dim}M = 1 \\ \|\psi\| = 1}} \langle \psi, A_W\psi\rangle \\ &\overset{9}{=} \mu_1(A_W). \end{align}

4. Equality (9) by p.1 and 3
5. Equality (2) by p.2.
6. By p. 2 $\mu_1(A_W)$ is eigenvalue of $A.$
7. Similar to inequality 4, $E \ge \mu_1(A).$
8. Assumption $E > \mu_2(A) \ge \mu_1(A)$ leads to contradiction at equality (1).
9. Eq. (6): $\inf_{M_1\cup M_2} = \inf_{M_1}$ if $\inf_{M_2} \ge \inf_{M_1}$ and $M_1 \neq \emptyset.$