Norm of a Bounded Linear Operator

• Dec 30th 2008, 09:19 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 $\displaystyle B$ is bounded we define its norm by the equality

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

where $\displaystyle X$ 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

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

and why can I make this assertion? Thank you so much in advance, and happy new year!
• Dec 31st 2008, 02:26 AM
HTale
OK, so here's my thinking so far; by definition, we measure the operator norm $\displaystyle ||B||$ in such a way that we take the smallest such $\displaystyle C$ such that the operator $\displaystyle B$ is still bounded. But that is the same as saying, we pick the maximal $\displaystyle ||Bx||$ such that for any $\displaystyle C$, we have it is bounded, and hence our equality. Am I on the right lines? How do I put this all formally?
• Dec 31st 2008, 09:24 AM
Opalg
Quote:

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

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

where $\displaystyle X$ 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

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

and why can I make this assertion?

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

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

For the reverse inequality, note that if $\displaystyle C>\|B\|_i$ then $\displaystyle \|Bx\|\leqslant C\|x\|$ for all x. In particular, $\displaystyle \|Bx\|\leqslant C$ for all x with $\displaystyle \|x\|=1$ and so $\displaystyle \|B\|_s\leqslant C$. Taking the inf over all such C, you see that $\displaystyle \|B\|_s\leqslant\|B\|_i$.
• Jan 1st 2009, 06:39 AM
HTale
Thank you Opalg, that explains it perfectly.