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I'm not familiar with Putnam's book, but part (2) of this theorem should be available in any of the standard references, e.g., Conway's Course in Functional Analysis or Reed and Simon, Functional Anal …

Basically you're asking for the map to take almost every fiber into itself. For bounded operators this is equivalent to commuting with every operator of the form $M_g$ with $g \in L^\infty({\bf R}_+) …

If $H$ is finite dimensional there is a one-line solution: ${\rm tr}(JK - KJ) = 0$, so $JK - KJ$ cannot have positive spectrum. But it is false in general! Let $V$ be a partial isometry from $H$ onto …

Well, if $A$ is bounded and $B$ is compact then $AB$ is compact, so $AB$ cannot be the identity unless the Banach space is finite dimensional. Thus the left (or right) inverse of a compact operator on …

I think you want functions from $\mathbb{R}$ to $\mathbb{R}$, not to $\mathbb{R} \cup \{\infty\}$. Because if $f$ takes the value $\infty$ at $\lambda$ then $f(A)$ won't be self adjoint --- unless $\l …

I think the best way to answer this question is to direct you to an introductory textbook on the mathematics of quantum mechanics. One book I really like is Hilbert Space Operators in Quantum Physics …

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{ …

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 …