The set $B(X, Y )$ is a normed linear space with the operator norm.
If $T ∈ B(X, Y)$ and $S ∈ B(Y, Z)$ for $X, Y , Z$ normed linear spaces, then the composition
$ST ∈ B(X, Z)$ and $\|ST\| ≤ \|S\| \|T\|.$
I don't know how to prove this.
Here is my trial.
Since the composition if exist of 2 linear operators is a linear operator then $ST$ is a linear operator. To prove it is bounded we prove that $\|ST x\| < \infty$.
if $\|T\| = \sup_{\|x\|=1}\|Tx\| < \infty$.
and $\|S\| = \sup_{\|y\|=1}\|Sy\| < \infty$.
I don't know how to continue.
Best Answer
Let $0 \neq x \in X$. Then you have that $\Vert \frac{x}{\Vert x \Vert} \Vert = 1$. Thus we obtain $$\frac{1}{\Vert x \Vert} \Vert Tx \Vert = \Vert T\frac{x}{\Vert x \Vert} \Vert \leq \Vert T \Vert.$$ This implies shows that $\Vert Tx \Vert \leq \Vert T \Vert \Vert x \Vert$ for all $x \in X$. (Note: For $x = 0$ this is trivial.) Now from this it follows that $$ \Vert STx \Vert \leq \Vert S \Vert \Vert T x \Vert \leq \Vert S \Vert \Vert T \Vert \Vert x \Vert \qquad \text{for all } x \in X.$$ Finally, this implies $$ \Vert ST \Vert = \sup_{\Vert x \Vert = 1} \Vert STx \Vert \leq \sup_{\Vert x \Vert = 1} \Vert S \Vert \Vert T \Vert \Vert x \Vert = \Vert S \Vert \Vert T \Vert$$ and you are since the equality above implies $\Vert ST \Vert \leq \Vert S \Vert \Vert T \Vert < \infty$.