# Norms of compositions of bounded operators between different spaces

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...

$$\| AB \|_{\mathcal L(X,Z)} = \sup_{x\in X, \|x\|_X = 1} \|ABx\|_Z \leq \sup_{x\in X, \|x\|_X = 1} \|A\|_{\mathcal L(Y,Z)} \|Bx\|_Y = \|A\|_{\mathcal L(Y,Z)} \sup_{x\in X, \|x\|_X = 1} \|Bx\|_Y = \|A\|_{\mathcal L(Y,Z)} \|B\|_{\mathcal L(X,Y)}.$$
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\|##.