# The operator norm of the composition of linear bounded operators between Banach spaces.

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 $$\frac{1}{\Vert x \Vert} \Vert Tx \Vert = \Vert T\frac{x}{\Vert x \Vert} \Vert \leq \Vert T \Vert.$$ 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 $$\Vert ST \Vert \leq \Vert S \Vert \Vert T \Vert < \infty$$.