Prove this integral equation has a unique solution? I recently asked this question How does one solve these self referential integral equation? and realized I didn't have a proof that this self referential equation had a unique solution. Can someone prove (or disprove) that:
\begin{equation}
\label{self ref 2}
     \tilde P(  \lambda) = A \int_{- \infty}^{\infty} (\tilde P (\lambda-x) \tilde P (  x) )^2d( x)
\end{equation}
(Where $A$ is a normalization constant for the distribution) has a unique solution?
 A: Ok, I have the following. Not a proper answer but I think it can give some insight maybe. By the way I'm using $P$ instead $\tilde P$ for typing sake.
Let's say we have two different solutions $P_1(\lambda)$ and $P_2(\lambda)$. We define the following by subtracting both of them.
$$\phi(\lambda)=P_1(\lambda)-P_2(\lambda) = A \int_{-\infty}^{\infty} P_1^2(\lambda-x)P_1^2(x)-P_2^2(\lambda-x)P_2^2(x) dx$$
$$\phi(\lambda)= A \int_{-\infty}^{\infty} \left[P_1(\lambda-x)P_1(x)+P_2(\lambda-x)P_2(x)\right]\left[P_1(\lambda-x)P_1(x)-P_2(\lambda-x)P_2(x)\right] dx$$
Applying Cauchy-Schwarz inequality:
$$|\phi(\lambda)|^2 \leq |A|^2 \left(\int_{-\infty}^{\infty} \left|P_1(\lambda-x)P_1(x)+P_2(\lambda-x)P_2(x)\right|^2 dx\right) \left(\int_{-\infty}^{\infty}\left|P_1(\lambda-x)P_1(x)-P_2(\lambda-x)P_2(x)\right|^2 \right) dx$$
Let's define the following non-negative function corresponding to the left parenthesis thing:
$$\psi(\lambda) = \int_{-\infty}^{\infty} \left|P_1(\lambda-x)P_1(x)+P_2(\lambda-x)P_2(x)\right|^2 dx$$
Then we have:
$$|\phi(\lambda)|^2 \leq |A|^2  \psi(\lambda) \left(\int_{-\infty}^{\infty}\left|P_1(\lambda-x)P_1(x)-P_2(\lambda-x)P_2(x)\right|^2 \right) dx$$
Integrating both sides w.r.t $\lambda$
$$\int_{-\infty}^{\infty}|\phi(\lambda)|^2 d\lambda \leq |A|^2 \int_{-\infty}^{\infty} \psi(\lambda) \left(\int_{-\infty}^{\infty}\left|P_1(\lambda-x)P_1(x)-P_2(\lambda-x)P_2(x)\right|^2 dx\right) d\lambda$$
working out the inside parenthesis:
$$\int_{-\infty}^{\infty}\left|P_1^2(\lambda-x)P_1^2(x)+P_2^2(\lambda-x)P_2^2(x) -2P_1(\lambda-x)P_1(x)P_2(\lambda-x)P_2(x)\right| dx$$
I have my doubts in the following but let's see anyway. If we can assume that $P_1(\lambda-x)P_1(x)P_2(\lambda-x)P_2(x) \leq P_2^2(\lambda-x)P_2^2(x)$ because $P$ is a probability distribution bounded by $0 \leq P \leq 1$, then:
$$\int_{-\infty}^{\infty}\left|P_1^2(\lambda-x)P_1^2(x)+P_2^2(\lambda-x)P_2^2(x) -2P_1(\lambda-x)P_1(x)P_2(\lambda-x)P_2(x)\right| dx$$
$$\leq \left|P_1(\lambda)+P_2(\lambda) -2 \int_{-\infty}^{\infty}P_2^2(\lambda-x)P_2^2(x) dx \right| = \left|P_1(\lambda)+P_2(\lambda) -2P_2(\lambda) \right| = \left|P_1(\lambda)-P_2(\lambda) \right|$$
This way we have:
$$\int_{-\infty}^{\infty}|\phi(\lambda)|^2 d\lambda \leq |A|^2 \int_{-\infty}^{\infty} \psi(\lambda) \left|P_1(\lambda)-P_2(\lambda) \right| d\lambda$$
$$\int_{-\infty}^{\infty}|\phi(\lambda)|^2 d\lambda \leq |A|^2 \int_{-\infty}^{\infty} \psi(\lambda) |\phi(\lambda)| d\lambda$$
$$\int_{-\infty}^{\infty}|\phi(\lambda)|^2 d\lambda - |A|^2 \int_{-\infty}^{\infty} \psi(\lambda) |\phi(\lambda)| d\lambda \leq 0$$
$$\int_{-\infty}^{\infty}|\phi(\lambda)|\left(|\phi(\lambda)| - |A|^2 \psi(\lambda)\right)  d\lambda \leq 0$$
So, at this point you either have:
$$|\phi(\lambda)| - |A|^2 \psi(\lambda) = 0$$
or
$$|\phi(\lambda)|=0$$
in which case would imply that $P_1(\lambda)=P_2(\lambda)$ and therefore unique solution to the original equation.
