An inequality proof in analysis

Question

Let $T \in L(X,Y), S \in L(Y, Z)$, where $X,Y,Z$ are normed vector spaces. Show that $\|S \circ T \| \leq \|S\|\|T\|$.

I tried to show that $\|\frac{S\circ T}{S}\|\leq \|T\|$, but I do not know how to evaluate $\|\frac{S\circ T}{S}\|$.

Thank you all.

Answer 1

Using a definition $$ \|S\|=\sup_{\|x\|=1}\|Sx\|=\sup_{x\ne 0}\frac{\|Sx\|}{\|x\|}, $$ and a property, derived by the definition $$ \|Sx\|\le \|S\|\|x\|, $$ one obtains that, for every $x\in X$, $$ \|STx\|=\|S(Tx)\|\le \|S\|\|Tx\|\le \|S\|\|T\|\|x\|. $$ Hence $$ \|ST\|=\sup_{\|x\|=1}\|STx\|\le \|S\|\|T\|. $$

Answer 2

See any book on Functional Analysis. In particular, you will understand that $||\frac{S\circ T}{S}||,$ generally, has no sense.