$p$-Norm of Block Diagonal Matrix

Question

Let $A\in \mathbb{K}^{r\times r}$, $B\in \mathbb{K}^{(n-r)\times (n-r)}$ and $C = \mathbb{K}^{n\times n}$ such that \begin{equation} C = \begin{pmatrix} A & 0_{r\times (n-r)}\\ 0_{(n-r)\times r} & B\\ \end{pmatrix}. \end{equation} I want to prove $||C||_p = \max\{||A||_p, ||B||_p\}$ for all induced $p$-matrix norms.

I already showed the direction "$\ge$" by writing \begin{equation} ||C||_p = \max_{||x||_p = 1} ||Cx||_p \ge \max_{\substack{||x||_p = 1\\ x_j=0\ \forall\ j>r}} ||Cx||_p = \max_{||x||_p = 1} ||Ax||_p = ||A||_p \end{equation} and the same for $B$, thus $||C||_p \ge \max\{||A||_p, ||B||_p\}$.

However, I do not see an easy approach to prove the direction "$\le$". Any help is appreciated.

Answer 1

Hint to get the other inequality: use $p$-th powers.

Let $x \in \mathbb{R}^r$ and $y \in \mathbb{R}^{n-r}$ with $\left\|\begin{bmatrix} x \\ y \end{bmatrix} \right\|_p^p = \|x\|_p^p + \|y\|_p^p=1$.

Then try to get an estimate on $\left\|C \begin{bmatrix} x \\ y \end{bmatrix} \right\|_p^p$

Spoiler (solution)

$$\left\|C \begin{bmatrix}x \\y\end{bmatrix}\right\|_p^p = \|Ax\|_p^p + \|By\|_p^p \le \|A\|_p^p \|x\|^p_p + \|B\|_p^p \|y\|^p_p \le \max\{\|A\|_p^p, \|B\|_p^p\} $$