A problem in Lp spaces .

1. Apr 28, 2008

mmzaj

Hi

i'm looking for some class of functions $$\phi(t)$$ that satisfy :

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

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

satisfy :

1 - $$\int ds =1$$

2 - $$\left\|t\right\|_n=\left\|t\right\|_1 ; n=2,3,4 .....$$

obviously this problem is appropriately studied in $$L^p$$ spaces .
if i'm not mistaken , can you help me please , and if not , would you please advise .

Last edited: Apr 28, 2008
2. Apr 28, 2008

Pere Callahan

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?

3. Apr 28, 2008

mmzaj

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 $$\phi(t)$$ in :

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

this is an analysis problem , not else !!! but it would be nice if you help .

4. Apr 28, 2008

Pere Callahan

I can't because I don't know the answer

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.

5. Apr 28, 2008

gel

It's not possible, unless you allow the Dirac delta function $\phi(t)=\delta(t-t_0)$.
You can show that $\phi$ 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.

6. Apr 28, 2008

gel

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

I got that example by letting $\phi$ be $\lambda$ times a standard normal density. So the left hand side is $\lambda E(e^T)=\lambda e^{1/2}$ and the rhs is 1, and $\lambda=e^{-1/2}$ gives the solution.

7. Apr 29, 2008

mmzaj

$$\int \ e^t \phi(t) \, dt =\int \sum^{\infty}_{n=0}\frac{t^n}{n!} \phi(t) \,dt$$ =$$=\sum^{\infty}_{n=0} \int \frac{t^n}{n!}\phi(t) \,dt =\sum^{\infty}_{n=0}\frac{1}{n!} \left( \int \ t \phi(t) \, dt \right)^n =\exp\int \ t \phi(t) \, dt$$

so you see , they are the same !!

Last edited: Apr 29, 2008
8. Apr 29, 2008

Pere Callahan

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.

9. Apr 29, 2008

mmzaj

why !!?? if you can go one way , certainly you can go all the way back !!

10. Apr 29, 2008

Pere Callahan

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?

11. Apr 29, 2008

mmzaj

alright, I'll indulge you there ,

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

the LHS =

$$\sum^{\infty}_{n=0} \int \frac{t^n}{n!}\phi(t) \,dt$$

the RHS =

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

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 :

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

implies :

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

QED

12. Apr 29, 2008

Pere Callahan

No, as I showed by my example above.

Another example:
Take

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
13. Apr 29, 2008

mmzaj

ok , i'll have to admit it , you got me there !!! but at least we can say :

if

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

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 :

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

is one possible solution . as far as i'm concerned this is good enough .

14. Apr 29, 2008

Pere Callahan

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

15. Apr 29, 2008

mmzaj

nice ... $$\delta(t-t_0)$$ clearly satisfies the conditions...... i understand the part where $$\int\phi(t)\ dt = 1$$ , but where did you get the condition where $$\phi(t) \geq 0$$ ??? ... also i understand the part where $$\sigma^2 = 0$$ , but is this enough to conclude that $$\phi(t) = \delta(t)$$ exclusively ?? i mean 0 and $$\delta(t)$$ are rather trivial !!! aren't they ??

Last edited: Apr 29, 2008
16. Apr 29, 2008

gel

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 $\phi$ is continuous it follows that $\phi\ge 0$ (**)

There's a few issues here though
(*) I'm assuming $\phi$ 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 $\phi$ 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 $\phi=0$ 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 $\phi=0$ is not a solution.

17. Apr 29, 2008

mmzaj

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 $$\phi(t) =0$$ , it's my bad , i didn't pay attention to the case where n=0 !!!

and of course thanks to you too pere .