Perturbation in C*-Algebra Let x be an element in a C*-algebra A, is it true that if x approximately commute with every element in A, then x is near the centre of A? More precisely, I want to know whether the following is true： Let x be an element in a C*-algebra A with norm 1. Then for any $\epsilon>0$, there exist a $\delta>0$ such that the following holds: $\forall y\in A,||y||=1,||xy-yx||<\delta\Rightarrow dist（x, Centre(A)）<\epsilon$. This is true if A is the matrix algebra, but I was wondering whether it can be generalized to any C*-algebra
 A: The present question seems very close to the following one, which has been well-studied in the literature:

Let $A$ be a C*-algebra. Does there exist a constant $K>0$ such that, for each $x\in A$, the distance from $x$ to $Z(A)$ is bounded above by
  $$ K \sup\{ \Vert xy-yx \Vert \;:\; y\in A, \Vert y\Vert \leq 1\} ?$$

A useful discussion of what is known can be found in Section 1 of 
[MR2274022]
R. J. Archbold, D. W. B. Somerset, Measuring noncommutativity in C*-algebras.
JFA 242 (2007) no. 1, 247--271
According to this paper, the answer turns out to be "yes" if $A$ is a von Neumann algebra, or one of several natural classes of C*-algebra; but the answer is "no" in general, by an example in
[MR0215111]
R. V. Kadison, E. C. Lance, J. R. Ringrose, Derivations and automorphisms of operator algebras. II.
JFA 1 1967 204–221. 
I haven't been able to read the Kadison-Lance-Ringrose paper, but a similar example seems to be discussed in Example 3.3 of
[MR0482236] 
R. J. Archbold, On the norm of an inner derivation of a C*-algebra.
Math. Proc. Cambridge Philos. Soc. 84 (1978), no. 2, 273–291.
A: In general, the answer to this has to be "no". To construct a counterexample, let $S$ be a dense subset of the unit interval (0,1], and let $\mathcal{H}=\ell^2(S)$ be the Hilbert space with orthonormal basis $(e_s)_{s\in S}$. For each positive integer $n$, let $\mathcal{H}_n$ be the subspace of $\mathcal{H}$ generated by linear combinations of $e_s$ over $1/(n+1) < s\le 1/n$. We can define a $C^*$ sub-algebra $\mathcal{A}$ of the bounded linear operators $B(\mathcal{H})$ as follows: $A\in B(\mathcal{H})$ is in $\mathcal{A}$ if and only if the following are satisfied


*

*$A$ maps $\mathcal{H}_n$ into $\mathcal{H}_n$.

*There exists a sequence of complex numbers $a_n$ such that, for each fixed $n$,
$$\Vert Ae_s-a_ne_s\Vert+\Vert A^*e_s-\bar a_ne_s\Vert\to0$$
as $s\to 1/n$.


It is easily seen that this is a sub-$C^*$-algebra. I'm thinking of $\mathcal{A}$ intuitively as being those elements of the direct product (sum?) $\prod_nB(\mathcal{H}_n)$ "joined continuously at the edges", and roughly as continuous functions $S\to\mathbb{C}$ but allowing some non-commutativity within the intervals $(1/(n+1),1/n]$.
If operator $A$ is in its center then it must restrict to the center of each $B(\mathcal{H}_n)$, so must be a constant on each of these. That is, $Ae_s=\lambda_se_s$ for some $\lambda_s\in\mathbb{C}$, which is independent of $s$ over $1/(n+1) < s\le 1/n$. Also, $s\mapsto\lambda_s$ must be continuous at $s=1/n$, from which we see that $s\mapsto\lambda_s$ is constant. So, $\mathcal{A}$ has trivial center.
Fixing a positive integer $m$, define an operator $A\in\mathcal{A}$ by $Ae_s=\min(ms,1)e_s$. Then, $\Vert A\Vert = 1$ and ${\rm dist}(A,{\rm Center}(\mathcal{A}))=1/2$. On the other hand, the restriction of $A$ to each $\mathcal{H}_n$ is within a distance $1/(2(m+1))$ from its center. This gives $\Vert AB - BA\Vert \le \Vert B\Vert / (m+1)$ for all $B\in\mathcal{A}$.
Choosing $m$ as large as we like, this contradicts your claim.
