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Norm of linear functional

  1. Jun 30, 2011 #1
    Hi everyone,

    I have been studying "Optimization by Vector Space Methods", written by David Luenberger and I am stuck in an obvious point at first glance. My problem is in page 105, where the norm of a linear functional is expressed in alternative ways. The definition for the norm of a linear functional f is (from now on ||y|| denotes the norm of y and |y| the absolute value of y):
    ||f|| = inf{M: |f(x)| <= M*||x|| for every x in X} where X is the vector space the functional is defined on.

    now this definition can be modified and give:
    ||f|| = inf{M: |f(x/||x||)| <= M for every x in X and x not zero} since f is linear which is equivalent to:
    ||f|| = sup(|f(x)|/||x||) for x not zero or
    ||f|| = sup(|f(x)|) for ||x|| = 1

    now in the book it states that also:
    ||f|| = sup(|f(x)|) for ||x|| <= 1
    which I can't compehend given the previous definition.

    any help is much appreciated,
    zok
     
  2. jcsd
  3. Jun 30, 2011 #2
  4. Jun 30, 2011 #3

    tiny-tim

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    welcome to pf!

    hi zok! welcome to pf! :smile:
    aren't they the same?

    eg if ||x|| = 1, then f(0.5x) = 0.5f(x) since f is linear, so |f(0.5x)| < |f(x)| ?
     
  5. Jun 30, 2011 #4
  6. Jun 30, 2011 #5

    Fredrik

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    Note that [itex]\big\{|f(x)|\, \big| \,\|x\|=1\big\}\subset \big\{|f(x)|\,\big|\,\|x\|\leq 1\big\}[/itex]. Let's call the first set A and the second set B. We have [itex]A\subset B[/itex]. This means that [itex]\sup A\leq\sup B[/itex]. On the other hand, you can easily show that an arbitrary member of B is smaller than sup A. This means that sup A is an upper bound of B, but sup B is the least upper bound, so [itex]\sup B\leq\sup A[/itex].

    (Click the quote button if you want to see how I did the math).

    You might also want to check out this thread.
     
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