Estimate the Cauchy integral for matrix-valued functions Recently, I became familiar with the concept of the "matrix function via Cauchy integral", i.e.,
$$f(A):=\frac{1}{2\pi i}\int_\varGamma f(z)(zI-A)^{-1} \mathrm{d}z$$
Furthermore, it can be shown that
$$\left\|\frac{1}{2\pi i}\int_\varGamma f(z)(zI-A)^{-1} \mathrm{d}z\right\| \leq\frac{1}{2\pi}\int_\varGamma \lvert f(z)\rvert \|(zI-A)^{-1} \|\mathrm{d}z$$ 
where $\|\cdot \|$ denotes the induced $2-$norm.
But I can not prove and understand the reason of the above inequality. Can you help me prove it?
I really appreciate the time you spend to read this note.
 A: This is not correctly stated: the last $dz$ needs to be $|dz|$.
$$\left\|\frac{1}{2\pi i}\int_\varGamma f(z)(zI-A)^{-1} \mathrm{d}z\right\| \leq\frac{1}{2\pi}\int_\varGamma \lvert f(z)\rvert \left\| (zI-A)^{-1} \right\| |\mathrm{d}z|$$
By parametrizing $\Gamma$, integration over  $\Gamma$ reduces to integration over some interval $[a,b]$. The desired inequality is a special case of the integral triangle inequality. Whenever $g:[a,b]\to X$ is an integrable function taking values in a normed space $X$, the inequality 
$$
\left\|\int_a^b  g(t)\,dt \right\|\le  \int_a^b\|g(t)\|\,dt \tag{2}
$$
holds. In your case, $g(t) = f(z)(zI-A)^{-1} \gamma'(t)$ where $\gamma$ is a parametrization of $\Gamma$.  
One way to show (2) is to start with the triangle inequality for finite sums: 
$$
\left\|\sum_k  g(t_k) \right\|\le  \sum_k \|g(t_k)\|
$$
and pass to the limit. 
Another is to let  $I=\int_a^b  g(t)\,dt$ and use a norming functional for $I$: a linear functional  $\varphi:X\to \mathbb C$ of norm $1$, such that $\varphi \left(I\right)=\left\|I \right\|$. Then 
$$
\|I\| = \varphi \left(I\right) = \int_a^b \varphi(g(t))\,dt \le \int_a^b \|g(t)\|\,dt
$$
the inequality being pointwise.
