I'm working through M. Schechter's 'Principles of Functional Analysis' and I'm working through a proof on page 136 that shows that the spectral radius $r_{\sigma} (T) $ of a bounded linear operator $T: X \to X$ on a complex Banach space $X$ (which is in this text defined to be $r_{\sigma}\left(T\right):=\inf_{n\in\mathbb{N}}\left\Vert T^{n}\right\Vert ^{\frac{1}{n}}$) is equal to $\sup_{\lambda \in \sigma (T)} |\lambda|$ and also equal to $\lim_{n \to \infty} \left\Vert T^n \right\Vert^{\frac{1}{n}},$ where $\sigma (T)$ is the spectrum of $T$.
As part of the proof, Schechter uses the following bound, which I wish to justify. If $C$ is a circle centred at the origin of radius $a$,
$$\left\Vert \frac{1}{2\pi i}\oint_{C}z^{n}\left(zI-T\right)^{-1}dz\right\Vert \leq \frac{1}{2 \pi} a^n \max_C \left\Vert \left( zI - T \right)^{-1}\right\Vert \left(2 \pi a \right) = \left\Vert \left( zI - T \right)^{-1}\right\Vert a^{n+1}.$$
I'm struggling how to do this. My first thoughts from my very hazy recollection of complex analysis were that
$$\left\Vert \frac{1}{2\pi i}\oint_{C}z^{n}\left(zI-T\right)^{-1}dz\right\Vert \leq \frac{1}{2\pi} \oint _C\left\Vert z^n \left(zI -T \right)^{-1} \right\Vert dz = \frac{1}{2\pi} \oint _C\left\Vert z^n \right\Vert \left\Vert \left(zI -T \right)^{-1} \right\Vert dz$$
and perhaps
$$\frac{1}{2\pi} \oint _C\left\Vert z^n \right\Vert \left\Vert \left(zI -T \right)^{-1} \right\Vert dz \leq \frac{1}{2 \pi} \max_C \left\Vert \left( zI - T \right)^{-1}\right\Vert \cdot (2 \pi a )\oint_C \left\Vert z^n \right\Vert dz$$
but I'm not sure if these steps are properly justified. I wonder if someone could very kindly help me out/put me right/help me finish the steps?
In general, the integral triangle inequality holds for functions $f$ from a measure space $\Omega$ to a normed linear space $X$. It says that $$\left\|\int_\Omega f \right\|\le \int_\Omega \|f\| \tag1$$ One can then follow up with the pointwise inequality $\|f\|\le \sup_\Omega \|f\|$ and conclude that $$\left\|\int_\Omega f \right\|\le \sup_\Omega \|f\| \cdot |\Omega| \tag2$$ where $|\Omega| $ stands for the measure of $\Omega$.
Here $\Omega$ is the circle of radius $a$, its total (linear) measure is the length $2\pi a$. Since $|z|=a$ on the circle, we have
$$\left\| z^n(zI-T)^{-1}\right\| = a^n \left\| (zI-T)^{-1}\right\| \tag3$$ Now use (3) in (2): $$\left\| \int_C z^n(zI-T)^{-1}\right\| \le 2\pi a \cdot a^n \max_C \left\| (zI-T)^{-1}\right\| $$ which is the claimed estimate.