I believe that the answer is affirmative and I would be grateful to any comments on my attempt (see below) of proving this.
Let $A$ be a C*-algebra and denote by $Z(A)$ the centre of $A$.
- First of all, it is clear that $Z(A)$ is a subalgebra of $A$.
- Furthermore, it is easy to see that $Z(A)$ is closed under the $*$-operation.
- All elements of $Z(A)$ satisfy the so called C*-identity, since $Z(A)$ is a subset of $A$.
- $Z(A)$ is norm-closed.
The only step that I don't feel comfortable with is step 4. Here is how I would attempt to prove it:
Take a sequence $\{c_n\}_{n\in \mathbb{N}}$ of elements from $Z(A)$ converging to $c\in A$. We want to show that $c\in Z(A)$.
Let $a\in A$ be arbitrary. Clearly $c_n a - a c_n = 0$ holds for all $n\in \mathbb{N}$. By assumption $\lim_{n\to \infty} ||c-c_n|| = 0$.
Now,
$|| ac-ca || = || ac-ca + c_n a - a c_n || = || a(c-c_n) + (c_n-c)a) || \leq 2 ||a|| \cdot ||c-c_n|| $
Therefore $||ac-ca||=0$ and hence $ac=ca$. Right?
(Somehow, the last two lines give me a bad feeling in my stomach that I can't explain.)