# Function of a completely continuous operator

I would be most thankful if you could help me with this question. If $A$ is a completely continuous Hermitian operator on a Hilbert space $H$, for what class of functions $f$ can one define a function of operator $A$, $f(A)$, such that the following equality holds $$f(A) = \sum_{i = 1}^{\infty} f(a_{i})P_{i} ?$$

The spectral decomposition of $A$ is $$A = \sum_{i = 1}^{\infty} a_{i}P_{i}$$ where $a_i$ are the eigenvalues of $A$ and $P_i$ are projectors on the corresponding eigenvectors.

For such functions $f$ does the following manipulation make sense if the operator $A$ is also positive? $$\sqrt{A} - I = \sum_{i = 1}^{\infty} (\sqrt{a_{i}} - 1)P_{i} = \sum_{i = 1}^{\infty}\left\{ \frac{1}{\pi}\int_{0}^{\infty}\left( \frac{1}{1 + x} - \frac{1}{a_{i} + x} \right)\sqrt{x}dx\right\}P_{i} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{\pi} \int_{0}^{\infty}\left(\frac{1}{I + x} - \frac{1}{A + x}\right)\sqrt{x}dx$$ where $I$ is the identity operator.

Thank you!

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You can define $f(A)$ for all continuous $f \colon \sigma(A) \to \mathbb{C}$. For a compact operator, the continuity of course is only a restriction in $0$. – Daniel Fischer Jul 5 '13 at 17:33
@Daniel Fischer Thank you! How does one prove, using continuous functional calculus, that $f(A) = \sum_{i = 1}^{\infty}f(a_{i})P_{i}$ if $f$ is continuous on $\sigma\left( A \right)$ and $A$ is completely continuous? Also, does my manipulation involving $\sqrt{A} - I$ make sense? – user22208 Jul 5 '13 at 18:08
Hmm, looking at Rudin (Functional Analysis, Chapter 12), it seems one can define it even for any bounded Borel function on the spectrum. That chapter treats normal operators that need not be compact, I think for the compact case many things can be greatly simplified. I forgot most of the stuff, though. I am doubtful about your integrals, because the spectrum of a compact operator is countable, so the Lebesgue measure doesn't seem right. With the right interpretation of integrals, and the correct measure on the spectrum, you can do such things. – Daniel Fischer Jul 5 '13 at 18:23