TITLE: Product of bounded linear operators
QUESTION [5 upvotes]: I have 2 bounded linear operators $T_1, T_2$ such that $T_2:X\rightarrow Y$ and $T_1:Y\rightarrow Z$. I know that, by boundedness, $||T_2(x)||\leq||T_2||\,||x||$ and using the norm of $T$ defined as
$$||T||=\sup_{x\in D(T), x\neq0} \dfrac{||Tx||}{||x||}$$
How do I prove that $||T_1T_2||\leq||T_1||\,||T_2||$?
I have tried this $||T_1T_2||=\sup_{x\in D(T),x\neq0}\dfrac{||T_1(T_2x)||}{||x||}$ and must now make use of boundedness, but am stuck.

REPLY [2 votes]: $$
\|T_1 T_2\| = \sup_{x \ne 0} \frac{\|T_1T_2x\|}{\|x\|} = \sup_{T_2x \ne 0} \frac{\|T_1T_2x\|}{\|T_2x\|} \frac{\|T_2x\|}{\|x\|} \le \sup_{y \ne 0} \frac{\|T_1y\|}{\|y\|} \sup_{x \ne 0} \frac{\|T_2x\|}{\|x\|} 
$$
Therefore:
$$
\|T_1 T_2\| \le \|T_1\| \|T_2\|
$$