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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 $ …

Yes, we have a spectral theorem for operators on real Hilbert spaces. The multiplication operator version says that there is a Hilbert space isomorphism between $H$ and some real $L^2$ space which tur …

A bounded operator on $X^*$ is the adjoint of a bounded operator on $X$ if and only if it is continuous for the weak* topology on $X^*$. This is still true if $X$ is a normed linear space, not necessa …

It's still true in the unbounded case, and you can see this using polar decomposition. Write $A = BU$ where $B$ is some positive unbounded operator on $K$ and $U$ is the orthogonal projection from $H$ …

Sure, let $A = (\bigoplus M_n ) + \mathbb{C}\cdot P$ where $P$ is a projection of the form $P = (p_n)$ with each $p_n \in M_n$ a rank 1 projection. This is a one-dimensional extension of $\bigoplus M_ …

What may be tripping you up here is that $T^+$, as you have defined it, could be unbounded, if the range of $T$ is not closed. So the answer to Is the formula (1) true only for matrix or even for …

No, let $\tau$ be integration against counting measure on $l^\infty$ and let $x_n$ be the characteristic function of $\{i: i \geq n\}$.

Yes, the unilateral shift $S$ on $l^2(\mathbb{N})$ generates $B(l^2(\mathbb{N}))$ as a von Neumann algebra. This is a consequence of the double commutant theorem and the fact that the only bounded ope …

No, this already fails for linear functionals. For instance, let $\zeta \in \beta \mathbb{D} \setminus \mathbb{D}$, where $\beta \mathbb{D}$ is the Stone-Cech compactification. Then $f \mapsto f(\zeta …

No, this does not exist, unless you allow the trivial solution $I = \{0\}$. Otherwise let $\tau$ be a tracial state on $I$ and let $(e_\lambda)$ be a quasi-central approximate unit for $I$. This means …

Christian Remling pointed out that a counterexample to the original form of the question is given by $F: [0,1] \to L(H, \mathbb{R}^1)$, $F(t) = t\langle \cdot, e_1\rangle$, $v(0) = e_1$. Any continuou …

No, of course not. Even if $A$ is bounded, this would imply that $i_\omega A$ is self-adjoint. So you only have a chance if $A$ and $i_\omega$ commute.

Another way to do it is to consider the bounded operators $(T_i + iI)^{-1}$, check that these are normal and commute so they can be simultaneously represented as multiplication operators on an $L^2$ s …

Well, the discrete version of this question (where $Tf(m) = \sum K(m,n)f(n)$ maps $l^\infty$ to $l^\infty$) isn't so hard. That should give you an intuition for why it's true. I don't see any slick wa …

It sounds like you want the Arveson spectrum.