An Integral Functional Equation 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?
 A: EDIT
1)
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 $g$.
2)
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 $[0,n]$. 
