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Bounded operator

  1. Aug 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Is the derivative operator [tex]D:L^2(0,1)\to L^2(0,1)[/tex] bounded? In other words, is there a c>0 such that for all [tex]f\in L^2(0,1)[/tex],
    [tex]\|Df\|\leq c\|f\|?[/tex]​


    2. Relevant equations
    For all [tex]f\in L^2(0,1)[/tex],
    [tex]\|f\| = \int_0^1 |f|^2\,dx.[/tex]​


    3. The attempt at a solution
    I'm pretty sure the answer is no. Here's my work:

    Suppose [tex]c^2>0[/tex] satisfies the above requirements. Define [tex]f(x)=e^{(c+1)x}[/tex]. Then
    [tex]\|Df\| = \int_0^1 (c+1)^2e^{2(c+1)x}\,dx = (c+1)^2\|f\| > c^2\|f\|.[/tex]​
    But this contradicts the fact that [tex]\|Df\|\leq c^2\|f\|.[/tex] Thus, D is unbounded. Q.E.D.

    Is this correct?
     
  2. jcsd
  3. Aug 24, 2009 #2

    Dick

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    You have the right kind of example but the logic is a little overcomplicated. If you define f(x)=e^(dx) then ||Df||=d^2||f||. At this point you are done since d^2 can be chosen as large as you want. So there can't be a c such that ||Df||<c||f|| for any f.
     
  4. Aug 24, 2009 #3

    No, just d||f|| ... provided d>0 ... but that is just as good.
     
  5. Aug 24, 2009 #4

    Dick

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    Right. There should have been a square root on the definition of ||f|| in the original post. Thanks for catching that.
     
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