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!
Best Answer
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...