# Explicit solution of an operator equation

Suppose we have an operator $$A:C([0,\infty )) \to C([0,\infty ))$$ defined as $$(Af)(x)=\int _0 ^ x \dfrac{f(y)}{\sqrt{x^2-y^2}}dy$$ Now I want to prove that for arbitrary $$g\in C([0,\infty ))$$ the explicit solution of an equation $$Af=g$$ is $$f(y)=\dfrac{2}{\pi}\dfrac{d}{dy}\int _0 ^y \dfrac {sg(s)}{\sqrt{y^2-s^2}}ds$$

I can prove that if we assume that $$g$$ is in $$C^1$$, but I have no idea what to do in the general case.

• Your inversion formula seems to be missing a factor of $\frac{2}{\pi}$. – ComplexYetTrivial Nov 18 at 16:20
• @ComplexYetTrivial I guess you're right – Fat ninja Nov 18 at 16:25
• Maybe $C^1$ denseness in $C$? – mbartczak Nov 18 at 22:00
• @mbartczak I'm pretty sure an operator $A^{-1}:g\mapsto f$ isn't continuous – Fat ninja Nov 18 at 22:20
• For completeness, can you write down the proof for $g \in C^1$? – астон вілла олоф мэллбэрг Nov 22 at 8:53