Weinstein's bound on eigenvalues of self-adjoint operators

Question

Consider a complex, separable Hilbert space $H$ and a (densely defined) self-adjoint operator $A: \mathcal D(A)\to H$. Assume that $A$ admits an orthonormal basis of eigenvectors $(\varphi_n)_{n\in \mathbb N}$ with $A\varphi_n=\lambda_n \varphi_n$. Next, define for $\psi\in \mathcal D(A^2)$ with $\|\psi\|=1$ the following quantities: $$ A_\psi:=\langle \psi,A\psi\rangle\tag 1$$

and $$\Delta_\psi A:=\sqrt{\langle \psi, (A-A_\psi)^2\psi\rangle} \tag 2\quad .$$

In Weinstein, D. H. "Modified ritz method." Proceedings of the National Academy of Sciences 20.9 (1934): 529-532. the author shows that $$ |\lambda^*-A_\psi|\leq \Delta_\psi A\tag 3 \quad .$$

with $\lambda^*\in \sigma_p(A)$ an eigenvalue of $A$ to be defined below. That is, the interval $[A_\psi-\Delta_\psi A\,,A_\psi +\Delta_\psi A]$ contains at least one eigenvalue.

But in doing so, he assumes that there exists an eigenvalue $\lambda^*$ of $A$ which is the closest to $A_\psi$, i.e. for which $(\lambda^*-A_\psi)^2\leq (\lambda_n-A_\psi)^2$ holds for all $n\in \mathbb N$.

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Question: What further conditions ensure the existence of such a $\lambda^*$? Or does it always exist? If not, what are examples of operators where it does not exist?

Answer 1

Assume the contrapositive that $$[A_\psi-\Delta_\psi A\,,A_\psi+\Delta_\psi A]\cap \sigma_p(A)=\emptyset\tag{1}$$

Since the $\varphi_n$ are an orthonormal basis, we can write $$ \psi=\sum_{n=1}^\infty a_n\varphi_n $$ with $\sum_{n=1}^\infty |a_n|^2=\|\psi\|^2=1$. There exists some $n_0$ such that $a_{n_0}\ne 0$. Since $\psi\in\mathcal{D}(A)$ we get $$ (\Delta_\psi A)^2=\langle\psi,(A-A_\psi)^2\psi\rangle=\|(A-A_\psi)\psi\|^2= \sum_{n=1}^\infty|a_n|^2|\lambda_n-A_\psi|^2\tag{2} $$ Note that from $(1)$ follows that $|\lambda_n-A_\psi|>\Delta_\psi A$ for all $n$. Hence $|a_n||\lambda_n-A_\psi|\ge(\Delta_\psi A) |a_n|$ for all $n$ so from $(2)$ $$ (\Delta_\psi A)^2=\sum_{n=1}^\infty|a_n|^2|\lambda_n-A_\psi|^2\ge(\Delta_\psi A)^2\sum_{n=1}|a_n|^2=(\Delta_\psi A)^2\cdot 1=(\Delta_\psi A)^2\tag{3} $$ From $(3)$ it follows we must have term-by-term equality $|a_n||\lambda_n-A_\psi|=(\Delta_\psi A) |a_n|$ for all $n$. But $a_{n_0}\ne 0$ and $|\lambda_n-A_\psi|>\Delta_\psi A$ for all $n$, so $|a_{n_0}||\lambda_{n_0}-A_\psi|>(\Delta_\psi A) |a_{n_0}|$. This is a contradiction. Conclusion - $(1)$ is false.$\Box$

Edit: $(2)$ is based on the result that since $A$ is self-adjoint, if $\psi\in\mathcal{D}(A)$ then $\sum_{n=1}^\infty |\lambda_n a_n|^2<+\infty$ and $A\psi=\sum_{n=1}^\infty \lambda_n a_n \varphi_n$. See Brian C. Hall 'Quantum Theory for Mathematicians' Example 9.15.