# Norm of a Bounded Linear Operator

• December 30th 2008, 10:19 PM
HTale
Norm of a Bounded Linear Operator
Sorry to be asking so many questions! I'm confused by the definition of the norm of a bounded linear operator,

If $B$ is bounded we define its norm by the equality

$||B|| := \inf\{ C: ||Bx|| \leq C||x||, \forall x \in X \}$

where $X$ is a vector space. I don't quite understand how the infimum fits into all of this. Any explanations would be excellent. Then, related to that, how is this the case

$||B|| = \inf\{ C: ||Bx|| \leq C||x||, \forall x \,\, \text{s.t} \,\, ||x||\leq 1 \} = \sup\{||Bx|| : ||x||=1 \},$

and why can I make this assertion? Thank you so much in advance, and happy new year!
• December 31st 2008, 03:26 AM
HTale
OK, so here's my thinking so far; by definition, we measure the operator norm $||B||$ in such a way that we take the smallest such $C$ such that the operator $B$ is still bounded. But that is the same as saying, we pick the maximal $||Bx||$ such that for any $C$, we have it is bounded, and hence our equality. Am I on the right lines? How do I put this all formally?
• December 31st 2008, 10:24 AM
Opalg
Quote:

Originally Posted by HTale
If $B$ is bounded we define its norm by the equality

$||B|| := \inf\{ C: ||Bx|| \leq C||x||, \forall x \in X \}$

where $X$ is a vector space. I don't quite understand how the infimum fits into all of this. Any explanations would be excellent. Then, related to that, how is this the case

$||B|| = \inf\{ C: ||Bx|| \leq C||x||, \forall x \,\, \text{s.t} \,\, ||x||\leq 1 \} = \sup\{||Bx|| : ||x||=1 \},$

and why can I make this assertion?

Let's define a "sup" norm $\|B\|_s := \sup\{||Bx|| : ||x||=1 \}$, and an "inf" norm $||B||_i := \inf\{ C: ||Bx|| \leqslant C||x||, \forall x \in X \}$. We want to show that $\|B\|_s = \|B\|_i$.

First note that for any nonzero vector x, $\left\|\tfrac {\textstyle x}{\|x\|}\right\| = 1$. Therefore $\left\|B\tfrac {\textstyle x}{\|x\|}\right\|\leqslant \|B\|_s$. Multiply through by $\|x\|$ to see that $\|Bx\|\leqslant \|B\|_s\|x\|$ for all x. Therefore $\|B\|_s$ is a possible value for C in the definition of $\|B\|_i$. It follows that $\|B\|_i\leqslant\|B\|_s$.

For the reverse inequality, note that if $C>\|B\|_i$ then $\|Bx\|\leqslant C\|x\|$ for all x. In particular, $\|Bx\|\leqslant C$ for all x with $\|x\|=1$ and so $\|B\|_s\leqslant C$. Taking the inf over all such C, you see that $\|B\|_s\leqslant\|B\|_i$.
• January 1st 2009, 07:39 AM
HTale
Thank you Opalg, that explains it perfectly.