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Show that if a $C^*$-algebra $A$ is reflexive as a Banach space, then $A$ must be finite dimentional.

I tried to solve it; but, I could not. please help me for this exercise. Thanks a lot!

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closed as off-topic by Thomas, Davide Giraudo, user26857, Marconius, 6005 Nov 5 '15 at 3:11

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Lemma 0. Let maximal abelian $^*$-subalgebra $M$ of $C^*$ algebra $A$ which is finite dimensional, then (i) $M$ is a linear span of finite family of projections $\{p_1,\ldots,p_n\}\subset A$ such that $p_i p_j=0$. (ii) $A$ is finite dimensional.

See exercise 4.6.12 in Fundamentals of the theory of operator algebras. Vol.1. Elementary theory. R. V. Kadison, J. R. Ringrose.

Lemma 1. Any infinite dimensional $C^*$-algebra $A$ contains selfadjoint element with infinite spectrum.

Since $A$ is infinite dimensional by lemma 0, so does its maximal abelian $^*$-subalgebra $M$. Therefore there is infinite family of pairwise orthogonal projections $\{p_n:n\in\mathbb{N}\}$. Now define $a=\sum\nolimits_{n=1}^\infty 2^{-n} p_n$. Since $a p_k=2^{-k}p_k$ for all $k\in\mathbb{N}$, then $\{2^{-k}:k\in\mathbb{N}\}\subset \sigma(a)$. Hence $\sigma(a)$ is infinite. For details also see this discussion.

Lemma 2. Let $\Omega$ be a locally compact space, and $C_0(\Omega)$ is reflexive, then $\Omega$ is finite.

I'll borrow the main idea from this answer. Assume $\Omega$ is infinite, then we have infinite sequence of distinct point $(\omega_n)_{n=1}^\infty$. Consider sequence of linear functionals $\delta_n:C_0(\Omega)\to\mathbb{C}:f\mapsto f(\omega_n)$ of norm $1$ and an isometric operator $I:\ell_1\to C(\Omega)^*: a\mapsto \sum\nolimits_{n=1}^\infty a_n\delta_n$. Thus we may regard $\ell_1$ as closed subspace of $C(\Omega)^*$. Sicne $C_0(\Omega)$ is reflexive, then from this answer we know that so does $C(\Omega)^*$. Now we see that $\ell_1$ is reflexive as closed subspace of reflexive $C_0(\Omega)^*$. Contradiction, hence $\Omega$ is finite.

Proposition. Any reflxive $C^*$-algebra $A$ is finite dimensional.

Assume $A$ is infinite dimensional, then by lemma 1 we have a positive element $a\in A$ with infinite spectrum. Consider commutative $C^*$ subalgebra $C^*(a)$ generated by $a$. Since $C^*(a)$ is a closed subspace of reflxive space, then it is also reflexive as closed subspace of reflexive space. From this aswer we know that $C^*(a)\cong_1 C_0(\sigma(a)\setminus\{0\})$, hence $C_0(\sigma(a)\setminus\{0\})$ is reflexive. By lemma 2 we get that $\sigma(a)\setminus\{0\}$ is finite and so does $\sigma(a)$. Contradiction, hence $A$ is finite dimensional.

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  • $\begingroup$ Why is the maximal abelian *-subalgebra infinite? $\endgroup$ – math112358 Jan 5 at 5:42
  • $\begingroup$ @mathrookie, I fixed the statement of the lemma. Now it must be clear that $M$ is infinite-dimensional $\endgroup$ – Norbert Jan 6 at 13:53

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