does f(x) have unique fixed point?

105 Views Asked by At

Let $g$ be a probability density function. We can assume about $g$, whatever we like (Only important thing, we know about random variable Y,which has $g$ as p.d.f is $P(Y<0)>0$.) Next, let $V(x)$ be a continuous function, with property: $x+A<V(x)<x+B$. Fix $0<\gamma<1$ and $m>0$. Is there any chance to prove that

$$f(x)=\frac{\gamma}{m+1}\int_{0}^{\infty}V(y)g(y-x)dy$$

is contraction mapping?

I know, that for any $m$ and $\gamma$ there exist at least one fix point.But uniqueness of it, is something, I am looking for.