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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.

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

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