Ytukyg

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.

Let $0 neq x in X$. Then you have that $Vert frac{x}{Vert x Vert} Vert = 1$. Thus we obtain
$$frac{1}{Vert x Vert} Vert Tx Vert = Vert Tfrac{x}{Vert x Vert} Vert leq Vert T Vert.$$
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 $Vert ST Vert leq Vert S Vert Vert T Vert < infty$.