# Product of bounded linear operators

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.

Hint: $$\Vert T_1(T_2x)\Vert\leq\Vert T_1\Vert\Vert T_2(x)\Vert\leq\Vert T_1\Vert\Vert T_2\Vert\Vert x\Vert$$
$$\|T_1 T_2\| = \sup_{x \ne 0} \frac{\|T_1T_2x\|}{\|x\|} = \sup_{T_2x \ne 0} \frac{\|T_1T_2x\|}{\|T_2x\|} \frac{\|T_2x\|}{\|x\|} \le \sup_{y \ne 0} \frac{\|T_1y\|}{\|y\|} \sup_{x \ne 0} \frac{\|T_2x\|}{\|x\|}$$ Therefore: $$\|T_1 T_2\| \le \|T_1\| \|T_2\|$$