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Homework Help: Operator norm

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data

    http://img252.imageshack.us/img252/4844/56494936eo0.png [Broken]

    2. relevant equations
    BL = bounded linear space (or all operators which are bounded).

    3. The attempt at a solution

    I got for the first part:

    [tex]||A||_{BL} =||tf(t)||_{\infty} \leq ||f||_{\infty} [/tex] so

    [tex] ||A||_{BL} \leq 1[/tex]

    In second part I have to proof that: [tex] ||A||_{BL} \geq 1 [/tex] and thus the conclusion can be drawn.

    But I'm stuck at how to use the fact that E is banach. (At least that's the one thing I haven't used yet so I presume I need it for the second part)
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 9, 2008 #2

    Office_Shredder

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    The general strategy for this is to prove that ||A|| is bounded below the proposed value, then find a single example where the norm is greater than or equal to the proposed value.

    For example, if the map was f(t) -> 2f(t) we could show ||A||>=2 by citing the example f=1... |A(1)| = |2| = |2|*|1| and by definition this gives (assuming we've shown A is bounded) ||A||>=2.

    Usually it's just guesswork to find this and doesn't require extra information. In this case try f = t
     
  4. Dec 10, 2008 #3
    Yes that is clear but I'm worried that I might have to use the fact that E is banach. By the way you choose f(t)=t but why can't I choose for example f(t)=1 it is continuous and the supremum norm is one. Aren't there a lot possibilities for choosing this function?
     
  5. Dec 10, 2008 #4

    Office_Shredder

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    There are a lot of possibilities for the function you pick...

    The fact that E is Banach is unnecessary, but keep in mind though that you DID in fact use it when you chose not to prove that you're working over a normed vector space.
     
  6. Dec 10, 2008 #5
    If that's the case it would be unneccessary to put the exclamation sign there but my instructor DID so I'm probably missing something.
     
  7. Apr 3, 2011 #6
    what u write is wrong

    So,

    [tex]||A||_{BL} =sup{||A(f)||_{\infty} ,for ||f||_{\infty} \leq 1 [/tex]

    [tex]=sup{||tf(t)||_{\infty} ,for ||f||_{\infty} \leq 1 [/tex]

    [tex]=sup{|t|||f(t)||_{\infty} ,for ||f||_{\infty} \leq 1 [/tex]

    [tex]\leq sup{|t|||f|_{\infty}|t| ,for ||f||_{\infty} \leq 1 [/tex]

    [tex]\leq 1 [/tex]

    Now consider [tex] f^*(t)=-t+1 [/tex] for all t in [0,1] => [tex] ||f^*||_{\infty}=1 [/tex]

    notice that [tex] f^*(t)[/tex] is continuous in [0,1] and [tex] f(1)=0 [/tex]

    also

    [tex] 1=||A(f^*)||_{\infty} \leq ||A||_{BL}||f^*||_{\infty}=||A||_{BL}[/tex]


    enjoy
     
    Last edited by a moderator: May 5, 2017
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