Continuous vs holomorphic functional calculus

Question

Let $A$ be a bounded self-adjoint operator on a Hilbert space. Then, it has a continuous functional calculus, which I can think of as arising from the fact that polynomials are dense in $\mathcal{C}(\sigma(A))$. It also has a holomorphic functional calculus. Are these the same?

On the one hand, any holomorphic function on a neighborhood of $\sigma(A)$ is certainly continuous on $\sigma(A)$ and the two calculi agree in this case, thanks to Runge's theorem, as discussed at Why does the image of continuous functional calculus coincide with the holomorphic functional calculus when considering holomorphic functions?.

On the other hand, any uniform limit of polynomials (or any holomorphic functions) is holomorphic. This seems to suggest that the holomorphic and continuous functional calculi are valid for the exact same classes of functions. Is this correct?

Well, clearly it isn't, since there are plenty of continuous functions of one variable which aren't even differentiable, so they certainly don't admit holomorphic extensions on any neighborhood. I suspect that the issue in the last paragraph is that one needs to be careful with domains. Namely, maybe there is a sequence of polynomials which is uniformly convergent on $\sigma(A)$ but not on an open neighborhood thereof, and so the continuous functional calculus could apply while the holomorphic one does not?

Finally, if this is the correct reason why the continuous functional calculus is more general, then does this give an easy way to generalize the holomorphic calculus for any bounded operator? Namely, we get an isometry from the closure of $\mathcal{R}_A$ in $\mathcal{C}(\sigma(A))$ to $\mathcal{B}(V)$, where $\mathcal{R}_A$ is the space of rational functions whose poles are in $\rho(A)$, which by Runge's theorem accounts for the holomorphic functional calculus but again probably gets a bit more?

Thanks in advance!

Answer 1

Just to be clear: even on a fixed compact subset of $\mathbb R^2\approx \mathbb C$, most continuous functions cannot actually be well-approximated (meaning, for use, in sup norm) by holomorphic functions. After all, sup-norm-on-compact (that is, uniform-on-compact ...) limits of holomorphic functions are holomorphic.

So the "continuous" functional calculus is much more general than the "holomorphic" functional calculus, and cannot (so far as I know) be derived from the holo fun calc.

Although Runge's theorem is interesting, I think it is tangential to the genuine issues here. Somewhat similar to the operational fact that Egoroff's and Lusin's theorems about approximating (!?) (Borel-) measurable functions by continuous ones, while interesting and clarifying, turns out (!) by this year not to be the key idea...

Answer 2

Sorry again for the short/dismissive initial comment. I thought you were expecting limits of holomorphic functions to stay holomorphic in this context, so I jumped at that. I can't answer your final question, but I have some observations that might help.

To expand on Paul's note about Runge's theorem being a red-herring: Let's write $\mathcal{H}_A$ for the set of functions that are holomorphic on a neighborhood of $\sigma(A)$. Then clearly we have $\mathcal{R}_A \subset \mathcal{H}_A$, so $\overline{\mathcal{R}_A} \subset \overline{\mathcal{H}_A}$ (with the closure taken in $\mathcal{C}(\sigma(A))$). Runge's theorem (and the triangle inequality) tells you that the opposite inclusion is also true, i.e. $\overline{\mathcal{R}_A} = \overline{\mathcal{H}_A}$, since a limit point of $\mathcal{H}_A$ can be uniformly approximated by holomorphic functions, which can be uniformly approximated by rationals.

All of that is just to say: considering rational functions isn't really buying you anything. You might as well ask about the closure of the ordinary holomorphic calculus, since $\overline{\mathcal{R}_A} = \overline{\mathcal{H}_A}$. So I'll frame everything in terms of $\mathcal{H}_A$ from here on.

Depending on the choice of operator $A$, we might have $$\mathcal{H}_A \subsetneq \overline{\mathcal{H}_A} \subsetneq \mathcal{C}(\sigma(A)).$$

The first inclusion can be strict by the Weierstrass conversation in the comments. To see that the second inclusion may be strict, consider a shift operator $S$ with $S(e_n) = (e_{n+1})$ on a Hilbert space with basis $(e_n)_{n=1}^\infty$. You can show that $\sigma(S) = \overline{\mathbb{D}}$ (the closed unit disk in the complex plane). And, for example, $f(z) = \bar{z}$ is a continuous function which isn't the limit of holomorphics, by Paul's observation that a uniform limit of holomorphics would at least be holomorphic on $\mathbb{D}$ (the interior).

For normal operators, we know that the extension to $\overline{\mathcal{H}_A}$ will be just fine (but we might as well work with the continuous functional calculus at that point).

I suppose I'd reframe the question as: "For a non-normal operator $A$, does the holomorphic functional calculus extend to $\overline{\mathcal{H}_A} \rightarrow \mathcal{B}(\mathcal{V})$?" You might have some more luck searching for the closure of the holomorphic calculus in general rather than involving rationals.

My suspicion is that the answer to the extension question is yes but also that there's not much interest in the extra functions you might find in the closure. If the ordinary holomorphic functional calculus $\mathcal{H}_A \rightarrow \mathcal{B}(\mathcal{V})$ is continuous, then it should respect Cauchy sequences, and I would think that completeness of the latter space should be enough to guarantee a continuous extension. But maybe you're right to worry about the spectral mapping situation. I don't see any obvious examples of $A$ where $f \mapsto f(A)$ is unbounded, but maybe there's a clever example somewhere.