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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.

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Hint: $$ \Vert T_1(T_2x)\Vert\leq\Vert T_1\Vert\Vert T_2(x)\Vert\leq\Vert T_1\Vert\Vert T_2\Vert\Vert x\Vert $$

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

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    $\begingroup$ Bravo to you Bravo! $\endgroup$ – Vaolter Jul 19 at 11:25
  • $\begingroup$ Ahahaha the classic punchline $\endgroup$ – João Bravo Jul 19 at 14:26

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