# Proving the spectral radius formula using holomorphic functional calculus.

At the moment I am reading lecutre notes of a professor at my uni available online on his webpage here.

Now I want to understand the proof of the following theorem on page 222:

Theorem 5.20: Let $X$ be a nonzero complex Banach space and let $A \in L^c(X)$ (set of bounded linear operators form $X$ to $X$). Then the the spectrum of A, $\sigma(A)$, is non-empty and $$r_A := \lim_{n \to \infty}||A^n||^{1/n} = \sup_{\lambda \in \sigma(A)}|\lambda|.$$

The proof uses to my understanding what is called holomorphic functional calculus, which is introduced starting at page 213. My problems start at page 223 in the proof of Theorem 5.20. On the upper half of the page right after using the cauchy-integral formula, the author states that he uses the complex version of the Hahn-Banach Theorem to prove that $$\langle x^*, A^{n-1}x \rangle = ...=\frac{1}{2 \pi i}\int_{\gamma} \frac{\langle x^*,R(z)x \rangle}{z^{n+1}}dz$$ implies $$A^n= \frac{1}{2 \pi i}\int_\gamma \frac{R(z)}{z^{n+2}}dz= ...= \int_{0}^{1} \frac{R(\gamma(t))}{\gamma(t)^{n+1}}dt$$

But I do not see at all how this is possible.

Any suggestions or explanations are greatly appreciated. Thanks a lot in advance!

## 2 Answers

The formula $$\langle x^*, A^{n-1}x\rangle=\int_\gamma \left\langle x^*, \frac{R(z)}{2\pi i z^{n+1}}x\right\rangle$$ holds for all functionals $x^*$. Hahn Banach tells you that there are enough functionals to imply that this is an equality of operators: $$A^{n-1} = \frac1{2\pi i }\int_\gamma R(z)/z^{n+1}.$$ (The integral probably is supposed to converge in the weak or strong operator topology). This holds for any $n$, so the first equation of the second line follows. The second equation follows if you plug in the explicit form of $\gamma$.

• Shouldn't Hahn-Banach tell me somehow that by plugging in $\gamma$ in the first formula stated, that $$\frac{R(\gamma(t))}{\gamma (t)^{n}}x$$ is continuous on $[0,1]$ and then I can apply Lemma 5.7 on page 213 and the Definition 5.8 of the integral of Banach space valued functions on page 214? – vaoy Dec 30 '17 at 10:24
• The continuity of $t\mapsto\frac{R(\gamma(t))}{\gamma(t)^n}x$ follows from the continuity of $R(z)=z(\mathbb1 -z A)^{-1}$ on the disk of radius $1/r(A)$, the continuity of $1/\gamma(t)$ and the continuity of the "evaluation of at $x$" map ($B\mapsto Bx$) on space of bounded linear operators. Hahn-Banach is not needed for any of these statements. – s.harp Dec 30 '17 at 18:17
• Because of the continuity of that map the integral can be defined via Riemann sums and actually converges in the norm topology. – s.harp Dec 30 '17 at 18:24

The resolvent $R(\lambda)$ is a series in $1/\lambda$ for $|\lambda| > r_{\sigma}(A)$ that is given by $$R(\lambda) = \sum_{n=0}^{\infty}\frac{1}{\lambda^{n+1}}A^n.$$ (You can check that $AR(\lambda)-\lambda R(\lambda)$=I.) Because of this, $$\frac{1}{2\pi i} \oint_{|\lambda|=r > r_{\sigma}(A)}\lambda^n R(\lambda)d\lambda = A^n.$$ And, of course this makes sense in terms of the functional calculus. So something strikes me as wrong in what you've written.