Characteristic function integral

Question

Let $X$ be random variable and $\phi_X(t)$ its characteristic function, then if $$ \phi_X(u) = \int_0^1(\phi_X(ut))^2 dt, $$ implies $$ \int_0^1(\phi_X(ut))^2 dt = (\phi_X(u))^2 - u\ \phi_X^{(1)}(u), $$ where (1) means first derivative in $u$. I tried to use per partes, didn't help.

Answer 1

Note that $$ \phi_X(u) = \int_0^1\phi_X(ut)^2 dt\tag{$\circ$} $$ is equivalent, by the change of variable $t\to ut$, to $$ u\phi_X(u) = \int_0^u\phi_X(t)^2 dt $$ which, by differentiation, yields $$ u\phi'_X(u)+\phi_X(u) =\phi_X(u)^2 $$ Using once again $(\circ)$ to replace $\phi_X(u)$ in the LHS by the integral, this is equivalent to the desired identity.

Edit: In terms of random variables, the identity $(\circ)$ means that $$X=^dU\cdot(X_1+X_2)\tag{$\ast$}$$ where the random variables $(U,X_1,X_2)$ are independent, $U$ is uniform on $(0,1)$ and $X_1$ and $X_2$ are distributed like $X$.

Some solutions are the exponential distributions, since, if $X$ is exponential of parameter $a$, then $X_1+X_2$ is gamma $(2,a)$, and $U\cdot(X_1+X_2)$ is again exponential of parameter $a$. The goal of the exercise the question comes from, might be to show that, with the Dirac delta at $0$, these are the only distributions on the nonnegative real half line solving the identity $(\ast)$.