Why $A^*A =A$ implies that $A$ is a C$^*$ algebra (Proposition 5.2.8 of An Invitation to Quantum Groups and Duality by Thomas Timmermann)

Question

I am reading "An Invitation to Quantum Groups and Duality From Hopf Algebras to Multiplicative Unitaries and Beyond" by Thomas Timmermann. In the proposition 5.2.8 (page 117) the author provide a proof of the following fact:

Let $(A, \Delta)$ be a C$^*$-algebraic compact quantum group with Haar state $h$, and let $X$ be a unitary corepresentation operator of $(A, \Delta)$ on a Hilbert space $H$. Then

$$ C_X = \overline{span}\{[id \otimes h](X(id_{H} \otimes a)) : a \in A\}$$ is a C$^*$-algebra.

In the proof the author first shows that $[C_X^* C_X] = C_X$ and claims that this implies $C_X$ is a C$^*$-algebra.

In general if $A$ is a Banach subalgebra of $B(H)$, $A^*A = A$ does not implies that $A$ is a C$^*$-algebra, e.g. let

$$ A = \{ \begin{pmatrix} 0 & x\\ 0 & y\\ \end{pmatrix} \}.$$

I think I must miss something. Would anyone let me know why $[C_X^* C_X] = C_X$ implies $C_X$ is a C$^*$-algebra here. Thank you!

Answer 1

I assume that $A^* A$ denotes the algebra generated by product $a^* b$ with $a, b \in A$, but the following argument also applies to some others interpretation of $A^* A$.

If $A^* A =A$ then for any $a \in A$, $a$ can be approximated (in norm) by linear combination of elements of the form $a_1^* b_1 \dots a_n^* b_n$ but for such an element $c = a_1^* b_1 \dots a_n^* b_n$ one has $c^* = b_n ^* a_n \dots b_1^* a_1 \in A^* A$ hence is also in $A$. This proves that $a$ can be approximated by element whose adjoint is in $A$, hence $a^* \in A$.

As said in my comment your 'counterexample' only satisfies an inclusion $A^* A \subset A$.