# Norms of compositions of bounded operators between different spaces

1. Jan 26, 2014

### AxiomOfChoice

Suppose I have $B: X\to Y$ and $A: Y\to Z$, where $X,Y,Z$ are Banach spaces and $B\in \mathcal L(X,Y)$ and $A\in \mathcal L(Y,Z)$; that is, both of these operators are bounded. Does it follow that $AB \in \mathcal L(X,Z)$ and
$$\| AB \|_{\mathcal L(X,Z)} \leq \|A\|_{\mathcal L(Y,Z)} \|B\|_{\mathcal L(X,Y)}$$
It seems like this should be the case, but any time I try to prove a functional analytic result like this, I always get mired in uncertainty about the details...

2. Jan 26, 2014

### AxiomOfChoice

Hmmm...okay, well, I think I've found my answer in Eqn. 3.3 of this PDF. Does this suffice for a proof of this result:
$$\| AB \|_{\mathcal L(X,Z)} = \sup_{x\in X, \|x\|_X = 1} \|ABx\|_Z \leq \sup_{x\in X, \|x\|_X = 1} \|A\|_{\mathcal L(Y,Z)} \|Bx\|_Y = \|A\|_{\mathcal L(Y,Z)} \sup_{x\in X, \|x\|_X = 1} \|Bx\|_Y = \|A\|_{\mathcal L(Y,Z)} \|B\|_{\mathcal L(X,Y)}.$$

3. Jan 26, 2014

### WannabeNewton

This is ok. Another way of proving it is to note that $\|A\|$ is the smallest number $C$ such that $$\|Ax\|\leq \|A\|\|x\|$$ for each $x$ in the domain. It follows that for each $x\in X$: $$\|ABx\|\leq \|A\|\|Bx\|\leq \|A\|\|B\|\|x\|$$ So $\|A\|\|B\|$ is some number $C$ such that $\|ABx\|\leq C\|x\|$. It follows that $AB$ is bounded and that $\|AB\|\leq \|A\|\|B\|$, since $\|AB\|$ is by definition the smallest $C$ such that $\|ABx\|\leq C\|x\|$.