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Homework Help: Bounded linear operator

  1. Nov 22, 2008 #1
    1. The problem statement, all variables and given/known data

    http://img389.imageshack.us/img389/9272/33055553mf5.png [Broken]

    3. The attempt at a solution

    Via induction: for n=1 equality holds now assume that Vn=Jn.
    I introduce a dummy variable b and the fundamental theorem of calculus and change order of integration:

    [tex] V_{n+1}f(t) = \int_a^t \frac{(t-s)^n}{n!} f(s) \mbox{d}s = \int_a^t \frac{t-s}{n} \frac{(t-s)^{n-1} } {(n-1)!} \mbox{d}s = \int_a^t \int_s^t \frac{1}{n} \frac{(t-s)^{n-1} } {(n-1)!} f(s) \mbox{d}b\ \mbox{d}s = \int_a^t \frac{1}{n} \int_a^b \frac{(t-s)^{n-1} } {(n-1)!} f(s) \mbox{d}s\ \mbox{d}b = \int_a^t \frac{1}{n} J^n f(b) \mbox{d}b

    = \frac{1}{n} J(Jf) = \frac{1}{n} J^{n+1}f[/tex]

    But how do I get rid of the 1/n?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 22, 2008 #2
    First of all I would use a different "dummy" variable as [tex]b[/tex] is the right end of the interval. Let's us [tex]x[/tex].

    For the matter at hand, your fifth equality is wrong: you can't use the induction hypothesis here, since in order for it to work you have to have [tex]\left(x-s\right)^{n-1}[/tex] in the numerator and not [tex]\left(t-s\right)^{n-1}[/tex].

    I would take a slightly different approach (although induction is obviously the way to go).
  4. Nov 22, 2008 #3
    you should use integration by parts to integrate
    [tex] V_{n+1}f(t) = \int_a^t \frac{(t-s)^n}{n!} f(s) \mbox{d}s [/tex]
    try with
    [tex] \frac{(t-s)^n}{n!} = u , f(s)= dv [/tex]
    It works for me.
  5. Nov 22, 2008 #4
    Yes you're right, I should have used for example x. I still don't see the mistake it shouldn't read b-a it should read t-s.

    Which is?

    Yes that worked, thanks!
  6. Nov 23, 2008 #5
    I don't see that I need to convince you anymore seeing as you solved it, but the induction hypothesis reads


    Or, since the argument is now x and not t,


  7. Nov 23, 2008 #6
    Yes you're right I see my mistake now thanks.

    So I found this:

    V_{n+1}f = \int_a^t \frac{(t-s)^n}{n!}\ f(s)\ \mbox{d}s = \left[ Jf(s) \frac{(t-s)^n}{n!} \right] _a^t + \int_a^t \frac{(t-s)^{n-1}}{(n-1)!)}\ Jf(s) \mbox{d}s = V_n(Jf) = J^n (Jf) =J^{n+1}f
    Last edited: Nov 23, 2008
  8. Nov 23, 2008 #7
    Last edited by a moderator: May 3, 2017
  9. Nov 23, 2008 #8
    Show that for any function f mit sup norm 1, you have [itex]||J^nf||\leq||J^n\underline 1||[/itex] where [itex]\underline 1[/itex] is the constant one function. Calculate [itex]||J^n\underline 1||[/itex] and conclude!
  10. Nov 23, 2008 #9
    Suppose [itex]||J^nf||\leq||J^n\underline 1||[/itex] then how can I calculate [itex]||J^n\underline 1||[/itex]?

    Is it this:

    [tex] \int_a^b \int_a^t \left| \frac{(t-s)^{n-1}}{(n-1)!}\ \right| \mbox{d}s\ \mbox{d}t [/tex]
  11. Nov 23, 2008 #10
    What is the norm you are using? Normally, C([a,b]) is endowed with the sup norm. Are you using the L_1 norm?
  12. Nov 23, 2008 #11
    Yes I'm using the L_1 norm but what do you mean by 'C([a,b]) is endowed by the sup norm' that's only valid if we use the infinity norm, right?

    So you're saying that is should be:

    \left| \left| \int_a^t \frac{(t-s)^{n-1}}{(n-1)!}\ \mbox{d}s\ \right| \right| _{\infty}
  13. Nov 23, 2008 #12
    In your assignment it says "calculate the norm in the Banach space...". The space C[a,b] with the L^1 norm is not a Banach space, and so if you take this norm the space BL(C[a,b]) is not a Banach space either. This should imply that they meant the supremum norm (and in fact, like Pere correctly noted, unless explicitly said otherwise the space C[a,b] is taken with the supremum norm).
  14. Nov 23, 2008 #13
    So the expression below is wrong?


    \left| \left| \int_a^t \frac{(t-s)^{n-1}}{(n-1)!}\ \mbox{d}s\ \right| \right| _{\infty}


    Then what should it be?
    Last edited: Nov 23, 2008
  15. Nov 23, 2008 #14
    This is correct; you just need to evaluate the integral and take the [itex]\sup_{t\in[a,b]}[/itex]
  16. Nov 23, 2008 #15
    Indeed the answer is deduced from this expression but how do you prove that:

    ||J^nf||\leq||J^n\underline 1||
    [/itex] ?
  17. Nov 23, 2008 #16
    For any f with ||f||<=1, you can bound ||f(s)|| by 1, and you get what I wrote.
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