Determining whether an uncountable set of integral equations yield a unique solution I am interested in the set of numbers $\alpha>0$ for which there exists a function $g:\mathbb{R}\to[0,1]$ satisfying 
$$
 \forall r\in \mathbb{R} \qquad f(r) = \int\limits_\mathbb{R}\! g(\alpha r+u) \,d\mu(u)
$$ 
Where $f:\mathbb{R}\to[0,1]$ is a fixed known function, and $\mu$ is a fixed known probabiliy measure on $\mathbb{R}$. 
In particular, how can I tell if this set is a singleton?
 A: For simplicity, assume the probability measure can be represented by a density $p(u)$.  Then we want to find $g(x)$ to solve:
$f(r) = \int_{-\infty}^{\infty} g(\alpha r + u)p(u)du$   (for all $r \in \mathbb{R}$)
This reminds me of a convolution.  So define:
$y(x)=g(-x)$, with Laplace transform $Y(s)=\int_{-\infty}^{\infty} y(x)e^{-sx}dx = G(-s)$.
$q(x)=f(-x/\alpha)$, with Laplace transform $Q(s) = \alpha F(-\alpha s)$.  
Then your equation becomes:
$q(-\alpha r) = \int_{-\infty}^{\infty} y(-\alpha r-u)p(u)du$   (for all $r \in \mathbb{R}$)
So:
$q(x) = \int_{-\infty}^{\infty} y(x-u)p(u)du$  (for all $x \in \mathbb{R}$)
Taking the Laplace transform of both sides gives $Q(s) = Y(s)P(s)$, where $Q(s)$ and $P(s)$ are transforms of $q(x)$ and $p(x)$.  Thus:
$G(s) = \frac{\alpha F(\alpha s)}{P(-s)}$. 
This "typically" gives a unique solution $g(x)$ for any $\alpha>0$. Of course, you have to worry about details like regions of convergence, and zeros of $P(-s)$.  
Oh, I just noticed the extra condition $g(x) \in [0,1]$ for all $x\in\mathbb{R}$. So you would need an $\alpha$ such that the transform of $\alpha F(\alpha s)/P(-s)$ satisfies that condition...
