Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $T$ and $S$ be bounded linear operators on a Hilbert space $H$. Verify that: $||TS||\leq ||T||\cdot ||S||$.

The definition of the operator's norm is $||T||=\sup\{||Tv||_H: ||v||_H=1\}$.

Thanks for your help.

share|cite|improve this question
This is one of the most trivial properties of bounded linear operators in a normed space - in fact, it's true in many other norms than the one you presented. You should just think about it for a minute; the proof is a one-liner (often the knowledge that the proof is trivial is enough to help spur you along to the answer). – Taylor Martin Dec 2 '12 at 3:26
You are right. Thanks. – Hiperion Dec 2 '12 at 3:32
up vote 3 down vote accepted

Note that for any $x\in H$, $\|Sx\|\leq\|S\|\|x\|$ and similarly $\|Ty\|\leq\|T\|\|y\|$ for any $y\in H$. Thus we have, for any $v$ with $\|v\|=1$, $$ \|TSv\|\leq\|T\|\|Sv\|\leq\|T\|\|S\|\|v\|=\|T\|\|S\| $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.