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Unfortunately, no. Let $H_n$ be the orthogonal projection onto $\mathbb{C}\cdot e_n$. Then $H_n \to 0$ strongly, but the spectral projection $\chi_{\{1\}}(H_n)$ has dimension $1$ for all $n$. The seco …

Suppose the range of $E_A(\lambda)$ has strictly larger dimension than the range of $E_B(\lambda)$, for some $\lambda$. Then we can find a vector $v$ in the first range which is orthogonal to the seco …

Sure, let $\mu_n = \frac{1}{|n|+ 1}$, then for each $n$ the vector $\sqrt{\mu_{n-1}}e_n\oplus \pm\sqrt{\mu_n}e_{n-1}$ is an eigenvector with eigenvalue $\pm\sqrt{\mu_n\mu_{n-1}}$.

The condition $A^n \leq B^n$ for all $n$ defines the spectral order on the positive part of $B(H)$, usually written $A \preceq B$. It makes the positive part of any von Neumann algebra a complete latt …

Here is a simple example that shows that the idea of spectral theory on pre-Hilbert spaces in the sense you are asking is hopeless. Consider the pre-Hilbert space consisting of the restrictions of all …

I agree with user131781, but wanted to add that there is a strong form of the spectral theorem which requires separability. Namely: if $A \in B(H)$ is self-adjoint then there is a probability measure …

I agree with Andreas that the obvious straightforward interpretation of "the eigenvalues grow to infinity" is that the sequence of eigenvalues $(\lambda_n)$ increases to infinity. (And, counter to Sas …

Sure, for instance let $P$ be the orthogonal projection onto the closed span of the characteristic functions $\chi_{[n,n+1)}$ for $n \in \mathbb{N}$. You get property 1 because each of these functions …

Use the fact that $\sigma(ST) \cup \{0\} = \sigma(TS) \cup \{0\}$. So if $A$ and $B$ are positive then, except possibly for the point $0$, $\sigma(AB)$ equals $\sigma(A^{1/2}BA^{1/2})$. If $AB \neq 0$ …

You know the result when $f$ is continuous. Now what is the Borel functional calculus? Given any bounded linear map $T: V \to W$ between Banach spaces we have a second dual map $T^{**}: V^{**} \to W^{ …

My book Mathematical Quantization satisfies most of these conditions.

Any bounded Borel function $f: \mathbb{R} \to \mathbb{R}$. If $TS = ST$ then (taking adjoint of both sides) $S^*T = TS^*$. Therefore both ${\rm Re}(S) = \frac{1}{2}(S + S^*)$ and ${\rm Im}(S) = \frac{ …

Sure, why not? I think it's a neat idea! Probably there are lots of ways to do this. In the simplest case, where $C$ intersects the real line transversally at $a$ and $b$, I guess you could just let …

The "well-known" fact is, of course, false. According to a theorem of C. Apostol and B. Morrel (On uniform approximation of operators by simple models, Indiana Univ. Math. J. 26 (1977), 427–442), if $ …

I think most books start with the spectral theorem for a single bounded self-adjoint operator; Reed and Simon vol. 1 is a canonical example. But spectral theory for a single normal operator is essenti …