Almost commuting elements and functional calculus

Question

Let $A$ be an unitial $C^*$-algebra, $X \subseteq \mathbb{C}$ a compact set and $f : X \to \mathbb{C}$ a continuous function. Prove that for any $\epsilon > 0$ there is a $\delta > 0$ such that whenever $x,y \in A$ with $x$ normal, $\sigma (x) \subseteq X$, $\|y\| \leq 1$ and $\| xy-yx \| < \delta$, then $\|f(x)y - yf(x) \| < \epsilon$.

Proving this for $y$ self adjoint or $x$ positive, is very easy using approximation by polynomials, but I can’t seem to pass to the general case. The problem seems to be bounding the commutator of $x^*$ and $y$. I’ve tried writing $x$ or $y$ as a linear combination of self adjoint elements or positive elements but I can’t seem to find a good relation that would bound the commutator.

Answer 1

It seems you only need the following result:

Proposition: Let $R > 0$. Then, for any $\epsilon > 0$, there exists $\delta > 0$ (which can depend on $R$) s.t. for any normal element $x$ and arbitrary element $y$ of any $C^\ast$-algebra $A$, if $\|x\| \leq R$, $\|y\| \leq 1$, $\|[x, y]\| < \delta$, then $\|[x^\ast, y]\| < \epsilon$.

Proof: Assume to the contrary. Then there exists $\epsilon > 0$, a sequence of $C^\ast$-algebras $A_n$, a sequence of normal elements $x_n \in A_n$, and a sequence of arbitrary elements $y_n \in A_n$, s.t.,

1. $\|x_n\| \leq R$, $\|y_n\| \leq 1$ for all $n$;
2. $\|[x_n, y_n]\| \to 0$;
3. $\|[x_n^\ast, y_n]\| \geq \epsilon$ for all $n$.

Pick any free ultrafilter $\mathcal{U}$ on $\mathbb{N}$. Take the ultraproduct $C^\ast$-algebra $A = \prod_{n \to \mathcal{U}} A_n$. Then we have $x = (x_n)_\mathcal{U} \in A$ is normal and $y = (y_n)_\mathcal{U} \in A$. Since $\|[x_n, y_n]\| \to 0$, we have $x$ and $y$ commute. However, as $\|[x_n^\ast, y_n]\| \geq \epsilon$ for all $n$, we have $\|[x^\ast, y]\| \geq \epsilon$, so $x^\ast$ and $y$ do not commute. This contradicts Fuglede's theorem. $\square$

Remark: The proof above is unfortunately highly non-constructive and does nothing to provide an explicit bound on $\|[x^\ast, y]\|$.

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Just for completeness, I'll also include the rest of the proof, granted the Proposition.

Theorem: Let $X \subseteq \mathbb{C}$ be a compact set, $f: X \to \mathbb{C}$ be a continuous function. Then, for any $\epsilon > 0$, there exists $\delta > 0$ (which can depend on $X$ and $f$) s.t. for any normal element $x$ and arbitrary element $y$ of any unital $C^\ast$-algebra $A$ (if $f(0) = 0$, then we can also remove the unitality assumption), if $\sigma(x) \subseteq X$, $\|y\| \leq 1$, $\|[x, y]\| < \delta$, then $\|[f(x), y]\| < \epsilon$.

Proof: Per Stone-Weierstrass, there exists a polynomial $p(z, \bar{z})$ s.t. $\|p - f\|_{C(X)} < \frac{\epsilon}{4}$ (and $p(0) = 0$ if $f(0) = 0$). Let $R = \max\{|z|: z \in X\}$. Using induction on the degree of $p$, it is easy to check that, for any $\epsilon > 0$, there exists $\delta' > 0$ (which can depend on $R$ and $p$) s.t. for any normal element $x$ and arbitrary element $y$ of any unital $C^\ast$-algebra $A$ (the unitality assumption may be removed if $p(0) = 0$), if $\|x\| \leq R$, $\|y\| \leq 1$, $\|[x, y]\| < \delta'$, $\|[x^\ast, y]\| < \delta'$, then $\|[p(x), y]\| < \frac{\epsilon}{2}$. When $x$ is normal and $\sigma(x) \subseteq X$, we then also have,

$$\|[f(x), y]\| \leq 2\|f(x) - p(x)\|\|y\| + \|[p(x), y]\| < \epsilon$$

The result now follows from the Proposition. $\square$