Let $f$ be a non-negative function supported and integrable on the positive real axis, such that $$\int_0^\infty f(x+y)p(y) dy = c[p] f(x), $$ where $c[p]$ a number (functional) dependent on function $p$.

1)Suppose the above equation holds for every non-negative function $p$ supported and integrable on the positive real axis. Is $f(x) = a\exp(-kx)$ for some non-negative $k$ and $a$?

2) Suppose the above equation holds for some non-negative function $p$ supported and integrable on the positive real axis. I suppose $f(x)$ is not necessarily an exponential function. How can we characterize $f$.

I tried using Laplace transform but got stuck. Similarly I tried using Fourier transform with the coefficient on the upper half complex plane. The form I obtained is $$\tilde f(k) \tilde p(-k) = c[p]\tilde f(k).$$ How do I proceed from here? Perhaps I should take a completely different route?




It is clear that for the equation to have a solution $f$, we must have $c[\delta_{a + b}] = c[\delta_{a}]c[\delta_{b}]$ for all $a,b > 0$, where $\delta_a$ is the Dirac measure in $a$. This means that we must have $c[\delta_x] = e^{kx}$ for some $k$, so (if $c$ is linear) we have $c[p] = \int_0^\infty p(y) e^{ky} dy$. But then the equation implies that $$ \int_0^\infty f(x+y) p(y) dy = \int_0^\infty f(x) p(y) e^{ky} dy. $$ If this holds for all $p$, it does indeed follow that $f(x + y) = f(x)e^{ky}$, so $f(x) = f(0)e^{kx}$.

So the answer to 1) appears to be yes, and the sign of $k$ is determined by the integrability of $f$.

Equivalently, we can start directly from the integral representation of a linear functional

$$c[p] = \int_0^\infty g(y)p(y)dy, \,\text{for some integrable }g(y),$$

same as in the solution for 2) below, so

$$ \int_0^\infty f(x + y) p(y) dy = \int_0^\infty f(x)g(y) p(y) dy, $$ for all $p$. We have

$$f(x+y)=f(x)g(y)$$ which gives the exponential function solution for both $f$ and $g$.


Certainly not a complete answer here, but let's assume that $c[p] = \int_0^\infty g(y)p(y) dy$ for some $g$. Then $$ \int_0^\infty f(x + y) p(y) dy = \int_0^\infty f(x)g(y) p(y) dy. $$ Now, it is only assumed that this holds for "some" $p$ --- if these happen to include for instance $\frac{1}{n}1_{(0,n)}$, this means that the average of $f$ over the interval $(x,x+n)$ equals $f(x)$ times the average of $g$ over $[0,n]$.

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  • $\begingroup$ Even under your added condition of $c[p]=1, \forall p$, constant $f$ is still an exponential function with the exponent coefficient $k=0$, which is covered by the conclusion of 1), and not, violating it as you claimed. Besides, I am concerned about a less stringent condition of $c$ being a functional dependent on $p$. $\endgroup$ – Hans Oct 17 '13 at 16:43
  • $\begingroup$ But maybe I'm misunderstanding your question (because I did not think of specifying $c$ as adding conditions, rather on the contrary). Do you intend to solve the equation for some given $c$, or for any $c$ (this was my initial interpretation), or find the $(c,f)$ that satisfies it? $\endgroup$ – svangen Oct 17 '13 at 17:02
  • $\begingroup$ The question says "where c[p] is a functional ...", so $c$ is A functional such that the first equation holds. The statement does not imply any functional $c$ would do. The statement does preclude the set of such $c$ being empty either. More directly, find $f$ such that there is some $c$ such that the first equation holds. Regarding your edited first comment, $c[\delta_z] = \exp(-kz)$ satisfies your condition $c[\delta_1]=c[\delta_{\frac{1}{2}}]$, or any similar condition. So it does not violate 1) and your conclusion does not hold. $\endgroup$ – Hans Oct 17 '13 at 17:24
  • $\begingroup$ Actually, your edited comment gives the proof for 1). $\endgroup$ – Hans Oct 17 '13 at 17:29
  • $\begingroup$ The point I was trying to make was simply that for most choices of $c$, the equation would not have any solution $f$ (which wasn't clear to me apriori the way you had formulated the question). But I think we agree now. Let me know if you want me to elaborate on some step of the argument in this edit. $\endgroup$ – svangen Oct 17 '13 at 17:33

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