First kind Volterra integral equation (almost) I am looking at the following integral equation $\forall x \in \left[0,M\right]$ (and $0 \leq z \leq x < M$) where
$M > 0$ is a constant:
$$
\int_{0}^{x}\frac{g(z)}{\sqrt{x^{2} - xz}}{\rm d}z = 0
$$
Where $g$ is a square integrable function with respect to a probability measure $\mu$.
It looks very much like an instance of first kind Volterra integral. However the kernel
$$
\operatorname{K}\left(x,z\right) =
\frac{1}{\sqrt{x^{2} - xz}}
$$
is not square integrable, which (as I read here ) is needed to treat this integral as Volterra type of integral.
(1) If the function $\operatorname{g}$ is continuous, I think one can argue that on $\left[0,M\right]$,  $\operatorname{g}$ must be zero because of the strict positivity of the kernel. However, the method that I used failed to conclude.
Q: is there a way to treat this problem using the theory from integral equations? Is it true statement (1)?
 A: I don't have time to flesh it out but maybe this will help?
Just fix $M = 1$ for now. Define
$$K(x,z) := \frac{1}{\sqrt{x^2 - xz}} \cdot \chi_{\{0 \leq z \leq x \leq 1\}}(x,z).$$
$$ A(z) := \int_z^1\frac{1}{\sqrt{x^2 - xz}} dx =^? 2 \text{sech}^{-1}(\sqrt{z})$$
Let
$$T(g)(x) := \langle g, K(x, \cdot)\rangle.$$
Question: Is $T(g) \in L^1([0,1])$ if $g \in C([0,1])$?
Check: By Holder's inequality we have
$$ \int_0^1 |T(g)(x)| dx = \int_0^1 |\langle g, K(x,\cdot)\rangle| dx \leq \int_0^1 \|g\|_\infty \|K(x,\cdot)\|_1 dx = \|g\|_\infty \int_0^1 \|K(x,\cdot)\|_1dx = 2 \|g\|_\infty < \infty.$$
So $T : C([0,1]) \rightarrow L^1([0,1])$. In particular, this means we can apply Fubini-Tonelli.
Question: Is $A \in L^2([0,1])$?
Check: Notice that
$$ \|A\|_2^2 = \int_0^1 |A(z)|^2 dz = \int_0^1 \left|\int_x^1 \frac{1}{\sqrt{x^2-xz}}dx  \right|^2 dz \\
= \int_0^1 |2 \text{sech}^{-1}(\sqrt{z})|^2 dz < \infty.$$
(Double check me here -- I'm trusting Wolfram).
Question: How do we solve $T(g) = 0$?
Check: Assume $T(g) = 0$, then
$$ 0 = \int_0^1 T(g)(x) dx = \int_0^1 \int_0^x \frac{g(z)}{\sqrt{x^2 - xz}}dzdx = \int_0^1 \int_z^1 \frac{g(z)}{\sqrt{x^2-xz}}dxdz = \int_0^1 g(z) A(z) dz \\
= \langle g, A \rangle.$$
I am not sure if there are any nice tricks at this point.
