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A problem in Lp spaces .

  1. Apr 28, 2008 #1

    i'm looking for some class of functions [tex]\phi(t) [/tex] that satisfy :

    [tex]\int_T \ t^n \phi(t) \, dt = \left( \int_T \ t \phi(t) \, dx \right)^n ; n=0,1,2,3 ...[/tex]

    from what i understand - if i'm not mistaken - the problem transforms to finding a set of measure spaces whose measure [tex]\ ds =\phi(t)dt[/tex] , and nth norm of [tex] t [/tex]
    [tex]\left\|t\right\|_n = \left( \int \ t^n\ ds \right)^\frac{1}{n} [/tex]

    satisfy :

    1 - [tex]\int ds =1[/tex]

    2 - [tex]\left\|t\right\|_n=\left\|t\right\|_1 ; n=2,3,4 ..... [/tex]

    obviously this problem is appropriately studied in [tex] L^p [/tex] spaces .
    if i'm not mistaken , can you help me please , and if not , would you please advise .
    Last edited: Apr 28, 2008
  2. jcsd
  3. Apr 28, 2008 #2
    Your two conditions seem to be mere reformulations of the original question.

    If there exist such function phi then they do not have a nice structure, for example if phi satisfies your equation than 2 phi does not, similar for the sum of two such phi's.
    Why would you think there exists a single such phi?
  4. Apr 28, 2008 #3
    yes indeed , the two conditions are merely a reformulation , but a more formal one . i was hoping to transfer the problem to a field of math were it is more suitable to study . anyway , the problem is also equivalent to solving for [tex] \phi(t)[/tex] in :

    [tex]\int \ e^t \phi(t) \, dt = \exp\int \ t \phi(t) \, dt[/tex]

    this is an analysis problem , not else !!! but it would be nice if you help .
  5. Apr 28, 2008 #4
    I can't because I don't know the answer:smile:

    My feeling is there should be at least one such a function phi ... but I don't know right now how to prove it or find that function.
    Maybe you can introduce some artificial parameter, take some derivatives and transform the problem into a differential equation for phi ..a power series expansion for phi will not work because the integrals will not converge.
  6. Apr 28, 2008 #5


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    It's not possible, unless you allow the Dirac delta function [itex]\phi(t)=\delta(t-t_0)[/itex].
    You can show that [itex]\phi[/itex] would have to be non-negative with integral 1, so it defines a probability distribution, and would have variance 0. So it would have to describe a constant value.
  7. Apr 28, 2008 #6


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    They're not the same - this problem can be solved in lots of ways.
    For example, [itex]\phi(t)=\exp(-(t^2+1)/2)/\sqrt{2\pi}[/itex].

    I got that example by letting [itex]\phi[/itex] be [itex]\lambda[/itex] times a standard normal density. So the left hand side is [itex]\lambda E(e^T)=\lambda e^{1/2}[/itex] and the rhs is 1, and [itex]\lambda=e^{-1/2}[/itex] gives the solution.
  8. Apr 29, 2008 #7
    [tex]\int \ e^t \phi(t) \, dt =\int \sum^{\infty}_{n=0}\frac{t^n}{n!} \phi(t) \,dt[/tex] =[tex]=\sum^{\infty}_{n=0} \int \frac{t^n}{n!}\phi(t) \,dt =\sum^{\infty}_{n=0}\frac{1}{n!} \left( \int \ t \phi(t) \, dt \right)^n =\exp\int \ t \phi(t) \, dt[/tex]

    so you see , they are the same !!
    Last edited: Apr 29, 2008
  9. Apr 29, 2008 #8
    I think the two problem are not equivalent. From your original equaiton you can deduce the the equation involving exponentials but not the other way round.
  10. Apr 29, 2008 #9
    why !!?? if you can go one way , certainly you can go all the way back !!
  11. Apr 29, 2008 #10
    Hm think about this example. You know x=2, y=3 from which you can conclude x+y=5, how ever the informstion x+y=5 is not enough to recover x=2,y=3, would you agree?:smile:
  12. Apr 29, 2008 #11
    alright, I'll indulge you there ,

    [tex]\int \ e^t \phi(t) \, dt=\exp\int \ t \phi(t) \, dt[/tex]

    the LHS =

    [tex]\sum^{\infty}_{n=0} \int \frac{t^n}{n!}\phi(t) \,dt [/tex]

    the RHS =

    [tex]\sum^{\infty}_{n=0}\frac{1}{n!} \left( \int \ t \phi(t) \, dt \right)^n[/tex]

    now by equating both sides term by term we get :

    \int \ t^n \phi(t) \, dt = \left( \int \ t \phi(t) \, dx \right)^n ; n=0,1,2,3 ...

    now i have to justify the last step :

    consider this :

    \sum^{\infty}_{n=0} \frac{a_n}{n!} = \sum^{\infty}_{n=0} \frac{b_n}{n!}[/tex]

    implies :

    [tex]a_n = b_n \ , n=0,1,2 ....[/tex]

  13. Apr 29, 2008 #12
    No, as I showed by my example above.

    Another example:

    a_1 = 1, a_2 = 4, all other a_n =0
    b_1 = 2, b_2 = 2, all other b_n =0

    then both sums sum to 3 ...
    Last edited: Apr 29, 2008
  14. Apr 29, 2008 #13
    ok , i'll have to admit it , you got me there !!! but at least we can say :



    \sum^{\infty}_{n=0} \frac{a_n}{n!} = \sum^{\infty}_{n=0} \frac{b_n}{n!}

    then :

    a_n = b_n \ , n=0,1,2 ....

    is one solution . hence , if :

    \int \ e^t \phi(t) \, dt=\exp\int \ t \phi(t) \, dt

    then :


    \int \ t^n \phi(t) \, dt = \left( \int \ t \phi(t) \, dx \right)^n ; n=0,1,2,3 ...[/tex]

    is one possible solution . as far as i'm concerned this is good enough .
  15. Apr 29, 2008 #14
    Yes, it means exactly that you have the implication
    \int \ t^n \phi(t) \, dt = \left( \int \ t \phi(t) \, dx \right)^n ; n=0,1,2,3 .. \Rightarrow \int \ e^t \phi(t) \, dt=\exp\int \ t \phi(t) \, dt
    The converse is not true, as gel immediately noticed.

    This means if you find a solution for the left side of the impplication it will also be a soluton for the right hand side. It can however very well be that is no solution to the left hand side while the right hand side does allow a solution. So the two probleme are clearly not equivalent.

    Which of the two do you actually want to solve? In either way, gel gave the answer together with nice explanations, thanks gel:smile:
  16. Apr 29, 2008 #15
    nice ... [tex]\delta(t-t_0)[/tex] clearly satisfies the conditions...... i understand the part where [tex]\int\phi(t)\ dt = 1 [/tex] , but where did you get the condition where [tex]\phi(t) \geq 0[/tex] ??? ... also i understand the part where [tex] \sigma^2 = 0[/tex] , but is this enough to conclude that [tex]\phi(t) = \delta(t)[/tex] exclusively ?? i mean 0 and [tex]\delta(t) [/tex] are rather trivial !!! aren't they ??
    Last edited: Apr 29, 2008
  17. Apr 29, 2008 #16


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    If f(t)=tn, you have,
    \int f(t) \phi(t)\,dt = f\left(\int t\phi(t),dt\right).
    By linearity, this applies to any polynomial. So, replacing f by f2,
    \int f(t)^2 \phi(t)\,dt = f\left(\int t\phi(t),dt\right)^2\ge 0.
    Approximating any continuous function by polynomials, this extends to any continuous f (*). Assuming [itex]\phi[/itex] is continuous it follows that [itex]\phi\ge 0[/itex] (**)

    There's a few issues here though
    (*) I'm assuming [itex]\phi[/itex] goes to 0 fast enough at infinity, otherwise there could be problems taking this limit. Assuming exponential decay (or greater) as t->infinity is enough.
    (**) if [itex]\phi[/itex] is only assumed to be measurable, and not cts, then it will be >= 0 'almost everywhere', meaning everywhere except on a set of length 0.
    (***) the 'trivial' case [itex]\phi=0[/itex] doesn't make much sense because you get 00 on the rhs. If you adopt the convention 00=1, common when writing power series, polynomials etc, then [itex]\phi=0[/itex] is not a solution.
  18. Apr 29, 2008 #17
    thanks a lot , i have to express my admiration in your approach . i'll get to you later with some questions , but for now

    as for [tex]\phi(t) =0[/tex] , it's my bad , i didn't pay attention to the case where n=0 !!!

    and of course thanks to you too pere .
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