Convolution and analyticity Assume $f$ and $g$ are continuous and related to each other as
$$
f(x) = \int _{0}^{x-1} \Big ( (x- y)^2 - 1\Big )^{3/2}g(y) \, dy, \qquad x>1. 
$$
If we happen to know that $f$ is real analytic at some point $x_0$ can we deduce that there is a point $\phi (x_0)$ where $g$ is real analytic? (I suspect $\phi (x_0)$ should be one of the points $x_0 \pm 1$.)
I'm also very interested in incomplete answers such as possible ways to prove this.  
 A: The function $f$ is not analytic. Indeed, setting
\begin{eqnarray}
\eta(u_1,u_2)&=&\Big[(u_1-u_2)^2-1\Big]^{3/2},\\
\Psi(u_1,u_2)&=&\int_0^{u_2}\eta(u_1,t)g(t)\ dt,
\end{eqnarray}
we have
$$
f(x)=\Psi(x,x-1).
$$
It follows that
\begin{eqnarray}
f'(x)&=&\partial_1\Psi(x,x-1)+\partial_2\Psi(x,x-1)=\int_0^{x-1}\partial_1\eta(x,t)g(t)\ dt+\eta(x,x-1)g(x-1)\\
&=&\int_0^{x-1}\partial_1\eta(x,t)g(t)\ dt=3\int_0^{x-1}(x-t)\sqrt{(x-t)^2-1}\ g(t)\ dt;\\
f^{(2)}(x)&=&\int_0^{x-1}\partial_1^2\eta(x,t)g(t)\ dt+\partial_1\eta(x,x-1)g(x-1)\\
&=&\int_0^{x-1}\partial_1^2\eta(x,t)g(t)\ dt=3\int_0^{x-1}\frac{2(u_1-t)^2-1}{\sqrt{(u_1-t)^2-1}}g(t)\ dt.
\end{eqnarray}
Since
$$
\lim_{t \to x-1}|\partial_1^2\eta(x,t)g(t)|=\infty
$$
it is clear that $f^{(3)}(x)$ does not exist. Thus $f$ is not analytic.
A: Here is a start. the integral equation can be written as
$$  f(x+1) = \int _{0}^{x} \Big ( (x+1- t)^2 - 1\Big )^{3/2} g(t) \, dt$$
$$ = \int _{0}^{x} \Big( ((x-t)+1)^2 - 1\Big )^{3/2}g(t) \, dt, $$
$$\implies f(x+1) = \int _{0}^{x} \Big( (x-t)^2+2(x-t) \Big )^{3/2}g(t) \, dt, $$
Now, the right hand side is a convolution of two functions. You can try the laplace transform technique to solve the integral equations. Also, the above integral equation is known as Volterra integral equation( just let $f(x+1)=h(x)$ ) of the first kind. Solutions of some Volterra equations of the first kind can be found here. 
