What Functions in Lp Spaces Satisfy the Transformative Integral Equation?

[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 .

 

[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?

 

[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 .

 

[Pere Callahan]

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 .

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.

 

[gel]

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 ...

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.

 

[gel]

the problem is also equivalent to solving for \phi(t) in :

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

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.

 

[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 !

 

[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.

 

[mmzaj]

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

 

[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?

 

[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 :

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

now i have to justify the last step :

consider this :

<br /> \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

 

[Pere Callahan]

alright, I'll indulge you there ,
consider this :

<br /> \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

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 ...

 

[mmzaj]

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

if

<br /> <br /> \sum^{\infty}_{n=0} \frac{a_n}{n!} = \sum^{\infty}_{n=0} \frac{b_n}{n!}<br />

then :

<br /> a_n = b_n \ , n=0,1,2 ...<br />

is one solution . hence , if :

<br /> \int \ e^t \phi(t) \, dt=\exp\int \ t \phi(t) \, dt<br />

then :

<br /> <br /> \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 .

 

[Pere Callahan]

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

if

<br /> <br /> \sum^{\infty}_{n=0} \frac{a_n}{n!} = \sum^{\infty}_{n=0} \frac{b_n}{n!}<br />

then :

<br /> a_n = b_n \ , n=0,1,2 ...<br />

is one solution . hence , if :

<br /> \int \ e^t \phi(t) \, dt=\exp\int \ t \phi(t) \, dt<br />

then :

<br /> <br /> \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 .

Yes, it means exactly that you have the implication
<br /> \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<br />
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

 

[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 ??

 

[gel]

where did you get the condition where \phi(t) \geq 0 ?

If f(t)=tⁿ, you have,
<br /> \int f(t) \phi(t)\,dt = f\left(\int t\phi(t),dt\right).<br />
By linearity, this applies to any polynomial. So, replacing f by f²,
<br /> \int f(t)^2 \phi(t)\,dt = f\left(\int t\phi(t),dt\right)^2\ge 0.<br />
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.

 

[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

but is this enough to conclude that \phi(t) = \delta(t) exclusively ?? i mean 0 and \delta(t) are rather trivial ! aren't they ??

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 .