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Math Help - Norm of a Bounded Linear Operator

  1. #1
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    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!
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  2. #2
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    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?
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  3. #3
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    Quote Originally Posted by HTale View Post
    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.
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    Thank you Opalg, that explains it perfectly.
    Last edited by mr fantastic; January 1st 2009 at 01:14 PM.
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