Norms of compositions of bounded operators between different spaces

[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
<br /> \| AB \|_{\mathcal L(X,Z)} \leq \|A\|_{\mathcal L(Y,Z)} \|B\|_{\mathcal L(X,Y)}<br />
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...

 

[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:
<br /> \| 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)}.<br />

 

[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 $\|ABx\|\leq C\|x\|$.