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For $G\in C^1([0,1])$, the displayed expression solves $Tf=G$, where $T=H_{(0,1)}/i$ and $H_{(0,1)}$ is the finite Hilbert transform with denominator $t-x$. To derive it, put $S(z)=\sqrt{z(z-1)}\sim z$ at infinity, take $h$ to be the Cauchy transform of $i\rho G$, and set $w=h/S$. The Sokhotski–Plemelj formula gives $w_{+}+w_{-}=G$ and the displayed jump $f=w_{+}-w_{-}$. The endpoint behavior is at most inverse square root, and $w=O(z^{-2})$ at infinity. Cauchy reconstruction from a jump across a finite interval then gives $w=C[f]$ and hence $Tf=G$. Adding any constant multiple of $1/\rho$ gives another solution in the integrable endpoint class.

Multiplying an auxiliary sign-changing factor by a growing analytic factor can invalidate this reconstruction. For example, using $S$ instead of $1/S$ with $G=1$ produces the constant function $w=1/2$ and zero jump. The boundary sum alone is not enough: the entire remainder in reconstructing $w$ from its jump must also be controlled.

Orient the contour and call its left-hand side the plus side. At a smooth interior point $t$, take a local arclength parameter $\gamma (s)$ with $\gamma (0)=t$. The Cauchy principal value means the symmetric deletion limitThe two local pieces are deleted with the same arclength cutoff. For a sufficiently smooth regular contour, an equal-radius deletion about $t$ gives the same limit. The point is assumed to be away from endpoints and self-intersections.

To derive the Sokhotski–Plemelj formula, split the contour into a short smooth arc through $t$ and its complement. The complementary integral has the same limit from both sides. On the short arc subtract $f(z)$ from the numerator. Because $f$ is holomorphic nearby, $(f(\zeta )-f(z))/(\zeta -z)$ extends continuously as $\zeta$ approaches $z$; this part also has the same limit from either side. The remaining local term is $f(z)\int d\zeta /(\zeta -z)$. Straightening the oriented tangent shows its extra contribution is $+i\pi f(t)$ from the left and $-i\pi f(t)$ from the right: on a real segment this is the direct limit of ${\int }_{-\delta }^{\delta }du/(u\mp i\epsilon )$. Dividing by $2\pi i$ yieldsIn particular the jump is $w_{+}-w_{-}=f$, while the sum is the principal-value integral divided by $\pi i$. Smooth densities on a real interval satisfy the same formulas at every interior point, by the identical subtraction step with a Hölder-continuous remainder.

Now put $h=w/\tilde{w}$. On the interval, the sign reversal of the auxiliary boundary values givesThus the required Cauchy transform density isWhenever this density is integrable at the endpoints, taking $h=C[F]$ constructs a function with the required jump. Multiplication back gives $w_{+}+w_{-}=G$. The jump of the constructed $w$ isTo conclude that this jump solves the final singular equation, one must additionally justify $w=C[f]$. This reconstruction requires endpoint and infinity conditions; the sign reversal of $\tilde{w}$ alone does not ensure it. Here is a complete valid normalization. LetThe branch is holomorphic off $[0,1]$, with $S_{+}(x)=i\sqrt{x(1-x)}$ and $S_{-}=-S_{+}$. If $G$ extends smoothly to $[0,1]$, the density $F=G/{\tilde{w}}_{+}=i\sqrt{x(1-x)}G$ vanishes like a square root at each endpoint. Its Cauchy transform $h$ is bounded at the endpoints and is $O(z^{-1})$ at infinity. Therefore $w=h/S$ is at most inverse-square-root singular at each endpoint and is $O(z^{-2})$ at infinity. Its jump $f$ is integrable.

For Cauchy reconstruction from a jump across a finite interval, apply the Cauchy integral formula to $w$ on a large slit contour, and shrink its banks to $[0,1]$. The large circle contributes zero in the limit; each small endpoint circle contributes $O(\sqrt{\epsilon })$ and also vanishes. The two banks give the jump $f$, proving $w=C[f]$. The sum part of the Sokhotski–Plemelj formula now givesThis proves the intended finite Hilbert transform inversion with admissible endpoint behavior. Adding a multiple of $1/\sqrt{x(1-x)}$ to $f$ gives a homogeneous solution; equivalently one can add a constant multiple of $1/S$ to $w$, whose boundary sum is zero.

The omitted normalization is a real limitation of the printed claim. To see it explicitly, choose instead $\tilde{w}=S$ and $G=1$. This auxiliary function satisfies all the stated holomorphicity, nonvanishing and sign-reversal requirements. ButThe displayed integral identity follows from $\xi =(1+\cos \theta )/2$ and its continuation from real $z>1$. Hence the proposed jump formula gives $f=0$, which does not solve the equation with $G=1$. The failure is the nonzero constant at infinity: $w-C[f]=1/2$. Thus the final implication is valid with suitable endpoint and decay conditions, but is false for an unrestricted auxiliary function as printed. The valid reciprocal-square-root normalization above supplies an actual solution for $G$ smooth up to the two endpoints.