Looking for a book: $B(H)$ not reflexive

Question

I'm looking for a book with a proof that for an infinite dimensional Hilbert space, $B(H)$ is not reflexive. Thank you.

Answer 1

Clearly this is not answer you requested, but may be this will be helpful for someone else.

We may regard $c_0$ as a closed subspace of $\mathcal{B}(H)$. Indeed, take any orthnormal basis of vectors $(e_i)_{i\in I}\subset H$ and for each $s\in c_0(I)$ consider bounded linear operator $T_s:H\to H$ well defined by $T(e_i)=s_ie_i$. Consider map $T:c_0(I)\to\mathcal{B}(H):s\mapsto T_s$. It is an isometric embedding, because $\Vert T_s\Vert=\sup_{i\in I}|s_i|$. Therefore $c_0(I)$ is a closed subspace of $\mathcal{B}(H)$. Since $H$ is infinite dimensional, then $I$ is infinite. Take any countable $J\subset I$, then $c_0=c_0(J)$ is closed subspace of $c_0(I)$ and a fortiori of $\mathcal{B}(H)$. Closed subspace of reflexive space is reflexive, so if $\mathcal{B}(H)$ is reflexive, then so is $c_0$. But $c_0$ is not reflexive, so $\mathcal{B}(H)$ is not reflexive either.

Answer 2

There is a more general fact: A C*-algebra $A$ is reflexive if and only if it is finite-dimensional. In the infinite-dimensional case, there is a normal element $x\in A$ with infinite spectrum (actually, one can find a positive element with infinite spectrum). Therefore, by the spectral theorem, $C^*(x)$ is an infinite-dimensional commutative C*-algebra, hence of the form $C_0(X)$ for some infinite locally compact space $X$. Thus, there is a closed subspace of $C_0(X)$ (and in particular of $A$), isomorphic to $c_0$, hence $A$ is non-reflexive as this property passes to closed subspaces.

You will find other proofs of this fact in this thread.

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As for the reference, if $X$ is an infinite-dimensional Banach space with the approximation property, then $\mathscr{B}(X)$ is not reflexive. This is Theorem 4 on p. 247 in

J. Diestel and J. J. Uhl, Jr., Vector Measures, Providence, R.I., AMS, Mathematical Surveys, (15),‎ 1977.