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"Functional calculus" is a word used to describe the practice of taking some functions or formulas defined on complex numbers, and apply them in some way to certain kinds of operators, despite that operators are not complex numbers and so they are not in the domain of the function.

There are many kinds of functional calculus. (Correct me if I'm wrong about something: I'm studying these topics right now)

There is the so called Continuous functional calculus, that applies a continuous function defined on the spectrum of a normal operator in a unital C*algebra to that operator.

There is the Borel functional calculus that aims to apply a more general Borel function to a self-adjoint operator.

And there are the Riesz and Holomorphic functional calculus, that apply the analytic functions to some kind of operators.

I didn't catch the difference between these last two. Can someone please try to explain that to me briefly? Thank you.

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  • $\begingroup$ Did you want the difference between Holomorphic FC and Borel FC? Or what? I had a parsing error. $\endgroup$ – amcalde Aug 29 '14 at 13:56
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    $\begingroup$ Between Riesz functional calculus and Holomorphic FC. I often find these terms in simil contexts but here en.wikipedia.org/wiki/Continuous_functional_calculus in the last sentence seems that there is a difference between these two. $\endgroup$ – Benzio Aug 29 '14 at 14:16
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If you have a bounded operator $A$, then the holomorphic functional calculus is always an option, and it is based on Cauchy's integral representation: $$ f(A) = \frac{1}{2\pi i} \oint_{C} f(\lambda)``\frac{1}{\lambda I-A}"\,d\lambda = \frac{1}{2\pi i} \oint_{C} f(\lambda)(\lambda I-A)^{-1}\,d\lambda. $$ The contour $C$ is any simple closed rectifiable curve enclosing the spectrum $\sigma(A)$ in its interior, and $f$ is a holomorphic function on a neighborhood of $C$ and its interior. For all such functions $f$, $g$, $f(A)g(A)=(fg)(A)$. That is $f\mapsto f(A)$ is multiplicative and linear. You cannot extend this calculus to functions $f$ which are continuous, at least not in general. However, if $A$ is normal or selfadjoint, then the calculus does extend to continuous functions.

You can see why the holomorphic functional calculus fails to extend to general continuous functions: the resolvent $(\lambda I-A)^{-1}$ may have a pole of order $n > 1$. For example, if $N$ is a nilpotent matrix of order $n \ge 2$, then the resolvent is $$ \frac{1}{\lambda I - N} = \frac{1}{\lambda}\frac{1}{I-\frac{1}{\lambda}A}=\frac{1}{\lambda}\left[I+\frac{1}{\lambda}N+\frac{1}{\lambda^{2}}N^{2}+\cdots+\frac{1}{\lambda^{n-1}}N^{n-1}\right] $$ Then the holomorphic functional calculus gives $$ f(N) = \frac{f(0)}{0!}I+\frac{f'(0)}{1!}N+\cdots+\frac{f^{(n-1)}(0)}{(n-1)!}N^{n-1}. $$ So it is impossible for the functional calculus to extend to general continuous functions for such $N$. Nilpotent matrices arise naturally when studying even a general matrix, as can be seen from the Jordan Canonical form. In fact, you can derive the Jordan form from the holomorphic functional calculus. Other hard theorems become more or less obvious in this context too, such as the Cayley-Hamilton Theorem: If $p$ is the characteristic polynomial for a matrix $A$, then $p(\lambda)(\lambda I-A)^{-1}$ can be easily shown to have only removable singularities, which, by the Cauchy integral formula, gives $p(A)=0$.

So why are selfadjoint and normal operators different? Selfadjoint and normal $A$ are peculiar because you can show the following: $A^{2}x=0$ iff $Ax=0$. This forces the nilpotent terms to disappear. This is a big part of the reason why the functional calculus can extend to continuous functions.

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    $\begingroup$ Sorry, I know this is old, but could you please explain why your example shows you can't extend it to continuous functions? I'd really appreciate it. $\endgroup$ – Rick Sanchez Sep 16 '16 at 3:53
  • $\begingroup$ @RickSanchez : If you had a functional calculus that extended to continuous functions, then it would have to agree with the holomorphic calculus if there are reasonable continuity requirements, which would imply the existence of all orders of derivatives of the functions because of the necessity of expressions such as the above for various nilpotent operators, simply because of resolvents. $\endgroup$ – DisintegratingByParts Sep 16 '16 at 10:10
  • $\begingroup$ @DisintegratingByParts I have the same question as Rick Sanchez..you write: Then the holomorphic functional calculus gives $$ f(N) = \frac{f(0)}{0!}I+\frac{f'(0)}{1!}N+\cdots+\frac{f^{(n-1)}(0)}{(n-1)!}N^{n-1}. $$ I don't see how this relates to the Cauchy's integral representation you have above for example..where is the evaluation at $0$ coming from..could you give the explicit holomorphic functional calculus and explicit continuous functional calculus representations of f(N) and state why the continuous representation is invalid? I think this would clear up the confusion. $\endgroup$ – eurocoder Mar 13 '18 at 13:27

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