Does the Abel transform preserve analyticity?

Question

Let $I=(0,1]$ and $T=\{(x,y)\in I^2;x\geq y\}$. If functions $f:I\to\mathbb R$ and $w:T\to\mathbb R$ are analytic, is the function $A_wf:I\to\mathbb R$, $$ A_wf(y)=\int_y^1\frac{f(x)w(x,y)}{\sqrt{x^2-y^2}}dx, $$ analytic, at least on $(0,1)$? Is there a simple argument or a reference (or a counter example) for this?

By analytic I mean real-analytic, and at the boundary of a domain ($I$ or $T$) this means that the Taylor series converges to the function in a neighborhood of any boundary point intersected with the domain. Alternatively I could require that the functions can be extended to an analytic function in an open neighborhood of the domain. I am not so much interested in analyticity of $A_wf$ at $1$, so a possible counter example should be non-analytic in $(0,1)$.

The function $A_wf$ is the Abel transform of $f$ with a weight $w$. If the general case is too complicated, you can assume that $w\equiv1$. Notice that the weight $w$ is analytic (and thus bounded) at $x=y$ as well, so the only singularity comes from the square root.

Since all ingredients are analytic, it seems that $A_wf$ should be analytic, but I do not see how to turn this into a rigorous argument. Singularity of the integrand at $x=y$ makes differentiation of $A_wf$ inconvenient, so a brute force calculation looks like a messy path.

Answer 1

Here's a suggestion: Let $y_0$ be real. Then $w(z,w)$ exists as a holomorphic function on a complex polydisk neighborhood $U$ of the point $(y_0,y_0)$. Let $(y,y)$ be in this neighborhood, then define $$ A_wf(y)=\int_{y}^{y_0}\frac{f(x)w(x,y)}{\sqrt{x^2-y^2}}\,dx +\int_{y_0}^{1}\frac{f(x)w(x,y)}{\sqrt{x^2-y^2}}\,dx. $$ Here the integrals are understood as path intgerals along straight lines.The second integral is analytic in $y$. The first is, too, as you see when you replace $f(x)w(x,y)$ with its Taylor-series around the point $(y_0,y_0)$.