Induced norm problem

Question

For any matrix $A \in \mathbb{R}^{m \times n}$ prove:

$$\|A\|_{p, i}:=\sup _{u \neq 0} \frac{\|A u\|_p}{\|u\|_p}=\sup _{\|u\|_p=1}\|A u\|_p.$$

I found on the internet that

$$\|Au\|_{p, i} \leq \|A\|_{p, i}\|u\|_{p}$$

Can someone explain me why this property holds?

But isn't $$\|A\|_{p, i}:=\sup _{u \neq 0} \frac{\|A u\|_p}{\|u\|_p}$$ by definition of induced norm?

What should I prove there?

Also then $$\sup _{u \neq 0} \frac{\|A u\|_p}{\|u\|_p}=\sup _{\|u\|_p=1}\|A u\|_p.$$

We have

$$\sup _{u \neq 0} \frac{\|A u\|_p}{\|u\|_p}=\sup _{u \neq 0} \frac{\|A \frac{u}{\| u\|_p}\|_p}{\|\frac{u}{\| u\|_p}\|_p}=\sup _{u = 1} \frac{\|A u\|_p}{\|u\|_p}$$

Is this correct?

Answer 1

1. For any matrix $A \in \mathbb{R}^{m \times n}$ prove:

$$\|A\|_{p, i}:=\sup _{u \neq 0} \frac{\|A u\|_p}{\|u\|_p}=\sup _{\|u\|_p=1}\|A u\|_p.$$

Note that for any nonzero vector $u\in \mathbb{R}^n$, there exists some $v$ such that $u = \|u\|v$, where $\|v\| = 1$. Therefore, we have $$\|A\|_{p, i}:=\sup _{u \neq 0} \frac{\|A u\|_p}{\|u\|_p} =\sup _{\|v\|=1} \frac{\|A (\|u\|v)\|_p}{\|(\|u\|v)\|_p} = \sup _{\|v\|=1} \frac{\|A v\|_p}{\|v\|_p} = \sup _{\|v\|=1}\|Av\|_p.~\blacksquare$$

2. Show that

$$\|Au\|_{p, i} \leq \|A\|_{p, i}\|u\|_{p}.$$

Notice that for $u=0$, the result is trivially true. Now consider $u\neq 0$. Recall that $\|A\|_{p, i}:=\sup _{u \neq 0} \frac{\|A u\|_p}{\|u\|_p}$. Then for $$\|A_{p,i}\| = \sup _{u \neq 0} \frac{\|A u\|_p}{\|u\|_p} \geq \frac{\|A u\|_p}{\|u\|_p} \implies \|A u\|_p \leq \|A_{p,i}\|\|u\|_p.~\blacksquare$$

Answer 2

To prove that $||Ax||\leq ||A|| \text{ }||x||$ note that for $x\neq0$ you can write

$\frac{||Ax||}{||x||} = ||A\frac{x}{||x||}|| = ||Az||\leq ||A||$, where $z$ is a normalized vector, whose norm is 1. Therefore, the last inequality follows from the definition $||A||=\sup\{||Ax||: ||x||=1\}$.