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A discontinuous Functional

  1. Jan 8, 2008 #1
    1. The problem statement, all variables and given/known data

    I have to show that the functional [itex] C_n [/itex] on the space of polynomials on the interval [0,1], that takes the n'th coefficient ie

    [tex] C_n\left( \sum_{j=0}^m a_j t^j \right) = a_n [/tex]

    is discontinuous with respect to the supremum norm [itex] \|p\|_{\infty} = \sup_{t\in[0,1]}|p(t)| [/itex].

    2. Relevant equations

    A functional F on a normed space is continuous if and only if

    [tex] \sup_{\|x\|\leq 1} |F(x)| < \infty [/tex]


    3. The attempt at a solution

    Our normed space in the problem is the polynomials with the supremum norm.

    As a hint it says to consider

    [tex] p_k(t) = \frac{(1-t)^k}{bin(k,n)} [/tex]

    bin(k,n) is the binomial coefficient.

    The supremum of |p_k(t)| in the interval is attained when t = 0 so

    [tex] \|p_k\|_{\infty} = \frac{1}{bin(k,n)} [/tex]

    and I get by the binomial formula for (1-t)^k, that

    [tex] C_n(p_k) = (-1)^n \|p_k\|_{\infty} bin(k,n)[/tex]

    but then I get

    [tex] \sup_{\|p_k\|\leq 1} | \|p_k\|_{\infty} bin(k,n) | = bin(k,n) [/tex]

    but I think the point with hint was that this was supposed to be infinite so that C_n would be shown to be discontinuous. What am I doing wrong?
     
  2. jcsd
  3. Jan 8, 2008 #2

    Dick

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    I don't think there is anything wrong. You have pk->0 in your supremum norm as k->infinity (at least for n>0). If Cn were continuous this should mean Cn(pk)->0. But you have |Cn(pk)|=1. That's enough to show Cn is discontinuous without any explicit infinities anyplace, right?
     
  4. Jan 8, 2008 #3
    That actually sounds reasonable. But then my calculation of the functional norm ie

    [tex] \sup_{\|p_k\|\leq 1} | C_n(p_k) | [/tex]

    is wrong, since this has to be infinite if C_n has to be discontinuous, though I can't see the mistake.
    Maybe I'll just stick to argument that lim |C_n(p_k)| = 1 but lim p_k = 0, as you said. This shows |C_n| to be discont and therefore C_n is discont, because |.| is a cont map.
     
  5. Jan 8, 2008 #4

    Dick

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    You can do it that way as well. But then change pk to just be (1-t)^k. Now |pk|<=1. But Cn(pk)=bin(k,n) which is unbounded (for n>0).
     
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