Problem: "Let $(A,\|\cdot\|)$ be a $C^*$-algebra and let $B\subseteq A$ be a closed subalgebra such that $B$ is self-adjoint. Show that $(B,\|\cdot\|)$ is also a $C^*$-algebra."

I'll begin by recalling the following which I make use of and reference to in my attempt below:

  1. Saying that $B\subseteq A$ is self-adjoint means that $a\in B\iff a^*\in B$.

  2. A $C^*$-algebra is a Banach Algebra $(A,\|\cdot\|)$ together with an involution $a\mapsto a^*$ such that $\|a^*a\|=\|a\|^2$ for all $a\in A$ (the $C^*$-property).

  3. We call a map on an algebra $A$ is called an involution if (i) $(a^*)^*=a$, (ii) $(ab)^*=b^*a^*$ and (iii) $(\alpha a+\beta b)^*=\bar\alpha a^*+\bar\beta b^*$.

Attempt: In order to answer the question I need, in my mind, to do several things. Firstly, I need to verify that $(B,\|\cdot\|)$ is a Banach algebra. Secondly, I need to show that there is an involution on $B$ which additionally satisfies the $C^*$-property.

To see that $(B,\|\cdot\|)$ is a Banach algebra consider the following reasoning. We are given that $(A,\|\cdot\|)$ is a $C^*$-algebra, so that in particular, it is a Banach algebra. This necessitates that it is a Banach space. But $B$ is a closed subalgebra of $A$, in particular it is a closed linear subspace of the Banach space $(A,\|\cdot\|)$ and so must itself be a Banach space.

We next need to show that we have an involution on $B$ which satisfies the $C^*$-property, and here is where I am running into trouble.

The self-adjointness of $B$ induces a map $B\to B,\,a\mapsto a^*$. Let $b:=a^*\in B$ for some $a\in B$. Then, since $B$ is self-adjoint we have $b^*\in B$ and our map takes $b\mapsto b^*$, or equivalently that $a^*\mapsto(a^*)^*$. And here is my issue; how do I formally deduce that $(a^*)^*=a$? This is clearly what I am after, but what do I know about the $^*$-operation which lets me do this? It's the same problem I encounter for showing the other two criteria of an involution.

In my notes I have the following answer to the question "What other $C^*$-algebras are there?", in which it is stated that "In the general case, they are closed self-adjoint subalgebras of $(B(H),\|\cdot\|_{op})$". Does this mean that I should use what I know already about the adjoint of a bounded linear operator on a Hilbert space? Even though I am not given that this is the particular $C^*$-algebra that we are considering?


By assumption, $A$ is a $C^*$-algebra. You are defining the ${}^*$ operation on $B$ just using the same operation as in $A$. Since $A$ is a $C^*$-algebra, you already know that $(a^*)^*=a$ for all $a\in A$, and in particular this is true for all $a\in B$. Similarly, all the other properties of the involution for $B$ follow immediately from the fact that you know they are true for $A$.

  • $\begingroup$ I'm sorry for asking such a stupid question, I don't know why I completely missed that, thanks! $\endgroup$ – Jeremy Jeffrey James Mar 15 '18 at 22:45

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