How would you define the square of the linear operator

Question

If you define the linear operator norm of $A:X\to Y$ to be $$\|A\|_{op} = \inf\{C>0: \|Ax\|_Y \leq C\|x\|_X \text{ for all } x \in X \}$$

Then how would you define $\|A\|_{op}^2$?

My guess is you use the triangel inequality

$\|A\|_{op}^2 \leq \|A\|_{op}\|A\|_{op}$ but I'm not sure what that would mean for set. It would be $$\inf\{C>0: \|Ax\|_Y \leq C\|x\|_X\text{ for all }x\in X \}^2?$$ If so can someone give an interpretation of this?

Answer 1

I think what you want to consider is what might $\|A^2\|$ be, since $\|A\|^2$ is just the square of the norm we have found. Now, in general $\|A^2\|\le\|A\|^2$

But lets look at why this is:

$\|A^2\|=\sup_{0\ne x\in X}\frac{\|A^2(x)\|}{\|x\|}=\sup_{0\ne x\in X}\frac{\|A(A(x))\|}{\|x\|}\le\sup_{0\ne x\in X}\frac{\|A\|\|(A(x))\|}{\|x\|}\le \sup_{0\ne x\in X}\frac{\|A(A(x))\|}{\|x\|}$

$\le\sup_{0\ne x\in X}\frac{\|A\|\|A\|\|x\|}{\|x\|}=\|A\|^2 $

Thus $\|A^2\|\le\|A\|^2$.