# Inequality of bounded linear operators on Hilbert space

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.

-
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). –  Sargera Dec 2 '12 at 3:26
You are right. Thanks. –  Hiperion Dec 2 '12 at 3:32
add comment

## 1 Answer

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 $\|v\|=1$, $$\|TSv\|\leq\|T\|\|Sv\|\leq\|T\|\|S\|\|v\|=\|T\|\|S\|$$

-
add comment