# Reflexivity of a Banach space

I've run into a few problems in which reflexivity of a Banach space is given as a hypothesis. These problems are sometimes of the type where the banach space is specific/concrete, and sometimes it is just any Banach space. To me, reflexivity seems to be a hard condition to use. Is there an easy list of tricks to use this hypothesis? For example, does reflexivity imply any more tangible results via a standard theorem from functional analysis? An example of the type of answer I'm looking for is "Reflexivity often allows one to use the uniform boundedness principle" or "Reflexivity implies that the unit ball is weakly compact." But one of the reasons I'm asking these questions is because I feel I'm missing some other tricks that get used in functional analysis because I'm asked the question

"Show that every C* algebra that is reflexive as a Banach space is finite dimensional" and I feel that I simply don't know enough tricks/theorems to do this. (I could also use hints on this specific question, which might go some ways in revealing to me more tricks for reflexivity in general.)

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To give a brief description of how far I've gone in the suggested example question, I have thought "How can I relate C* algebras to anything in its Banach space dual? Oh, positive linear functionals are important... maybe I should use GNS." And then I'm stuck. – Jeff Oct 1 '12 at 1:20
Note that the proof you are looking for will imply in particular the fact that $\ell^\infty(\mathbb{N})$ and $C[0,1]$ are not reflexive. I don't think you can expect it to be too simple. – Martin Argerami Oct 3 '12 at 17:21
Interesting point. A friend advised me to show that every infinite dimensional C* algebra has a normal element of infinite spectrum, and then to go from there. I don't really know how to "go from there." – Jeff Oct 3 '12 at 18:17
It leads to the same idea I'm using in the answer. If you have a normal element of infinite spectrum, you can find infinitely many disjoint compact sets inside the spectrum; then using functional calculus you produce normal elements $b_1,b_2,\ldots$ with $b_ib_j=0$ if $i\ne j$. You represent your algebra faithfully on a Hilbert space, and in the closure (which is a von Neumann algebra) you can use Borel functional calculus on the $b_i's$ to get infinitely many pairwise orthogonal projections, and then you can construct the embedding of $\ell^\infty(\mathbb{N})$. – Martin Argerami Oct 3 '12 at 18:33
If a normal element $a$ has infinite spectrum $S$ you can find a convergent sequence $s_n \to s$ in $S$ with pairwise distinct $s_n \neq s$. As soon as you have that you can easily embed $c_0$ in $C(S)$ by choosing suitable disjointly supported bump functions $0 \leq f_n \leq 1$ such that $f_n(s_m) = \delta_{n,m}$ and sending $(x_n)$ to $\sum _n x_n \cdot f_n$. By the continuous functional calculus $C(S) = C^\ast(a)$, so $c_0$ embeds as a closed subspace into $A$, so $A$ cannot be reflexive since $c_0$ isn't reflexive. (@Martin maybe you're interested). – commenter Oct 7 '12 at 16:22

Suppose that $A$ is reflexive. By Sakai's theorem, $A$ is a von Neumann algebra, as it has a predual (being reflexive, $A$ is the dual of its dual). Being a von Neumann algebra, $A$ has abundance of projections, and in particular it has a maximal orthogonal set of projections $\{p_j\}_{j\in J}$. If $J$ is infinite, then there is a sequence $p_1,p_2,\ldots$ of pairwise orthogonal projections. This allows us to construct a copy of $\ell^\infty(\mathbb{N})$ inside $A$. As subspaces of reflexive Banach spaces are reflexive, we get a contradiction. This shows that every orthogonal family of projections in $A$ is finite, and this implies that $A$ is finite-dimensional.
Depends on what you call "elementary". I know about exercises in that book that are exactly the content of a paper. In any case, looking at the context of the question in the book, you can use the immediate previous question about masas to replace the first part of the proof (i.e. you only need to show that a masa contains a copy of $\ell^\infty(J)$, where the cardinality of $J$ agrees with the dimension of the masa). – Martin Argerami Oct 2 '12 at 14:05