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For continuous boundary data on the counterclockwise unit circle and its Cauchy transform $\Phi$, $\Phi (r\omega )-\Phi (r^{-1}\omega )=P\varphi (r\omega )$ for $0<r<1$. This follows by subtracting the two Cauchy kernels and obtaining the Poisson kernel on the circle. Consequently the paired interior-minus-exterior difference tends to $\varphi (\omega )$ as $r\uparrow 1$. For $r\downarrow 1$ from above, the same written difference has the opposite sign. This paired limit does not require each individual boundary value of the transform to exist.

// Destination: analysis.bigb

For $|\zeta |=1$ and $|z|<1$, the Poisson kernel on the circle satisfiesFor fixed $\zeta$, it is the real part of a holomorphic function of $z$, hence a harmonic function. On each compact subdisk, its derivatives are bounded uniformly in $\zeta$ because the denominator stays away from zero. Differentiation under the integral therefore proves that $P\varphi$ is harmonic, with the word interpreted componentwise when $\varphi$ is complex-valued.

The geometric seriesconverges uniformly in $\zeta$ for a fixed compact subdisk. With normalized Lebesgue measure, integration of its real part gives kernel mass one. Fix a boundary point $\omega$ and choose $\delta >0$ so that $|\varphi (\zeta )-\varphi (\omega )|<\epsilon$ when $|\zeta -\omega |<\delta$. ThenIf $z\to \omega$ from inside the unit disc, the remaining denominator is at least $(\delta /2{)}^2$ for sufficiently close $z$, while $1-|z{|}^2\to 0$. Hence $P\varphi (z)\to \varphi (\omega )$ along every interior approach, not just radial ones.

Orient the unit circle counterclockwise in the Cauchy transform. Its kernel and every $w$-derivative are uniformly bounded for $w$ in any compact subset of $\mathbb{C}\setminus \mathbb{T}$. Thus differentiation under the integral givesand proves that $\Phi$ is a holomorphic function on both components of its domain. Since $d\zeta =2\pi i\zeta dm(\zeta )$, for $0<r<1$ and $|\omega |=1$ the difference of its two kernels isConsequently the paired radial jump of a continuous Cauchy transform obeys the exact identityand the established Poisson integral boundary convergence givesThis proves the paired difference without assuming that the two individual boundary values exist for every continuous datum.

The printed direction $r\downarrow 1$ has the opposite sign. It is present in the original PDF, not merely in the converted TeX. With that direction $r>1$, so substituting $s=1/r<1$ in the exact identity givesThe simplest counterexample is $\varphi \equiv 1$: the Cauchy integral formula gives $\Phi (w)=1$ inside and $\Phi (w)=0$ outside, so the printed difference is constantly $-1$. The requested positive jump is corrected by taking $r\uparrow 1$, or by reversing the order of the difference when $r\downarrow 1$.

Orient the circle counterclockwise and define its Cauchy transformThe inside and outside limits satisfy ${\Phi }_{+}-{\Phi }_{-}=\varphi$ and ${\Phi }_{+}+{\Phi }_{-}=(\pi i{)}^{-1}\mathrm{PV}{\int }_C\varphi (\tau )/(\tau -t)d\tau$. Also ${\Phi }_{-}(z)=O(z^{-1})$ at infinity. Applying these formulas to the two polynomial coefficients and dividing by $t^2$ gives the Riemann–Hilbert conditionThe homogeneous jump is $G=t/(t^2-1)$. A canonical factorization of a rational scalar Riemann-Hilbert jump isThe inside factor is analytic and nonvanishing; the exterior factor has no zeros or poles outside the circle and grows like $z$. The jump index is $1-2=-1$, consistently with this growth.

Put $F_{+}={\Phi }_{+}$ and $F_{-}={\Phi }_{-}/X_{-}$. Their additive jump is $R(t)=((A-1)t+1)/(t^2-1)$. Since $R$ is analytic outside the circle and decays there, the functions $F_{+}$ inside and $F_{-}+R$ outside glue to an entire function. This function tends to zero at infinity, so Liouville's theorem gives $F_{+}=0$, $F_{-}=-R$. ConsequentlyThe required decay of a Cauchy transform forces $A=1$. Then ${\Phi }_{-}=-1/z$, and the jump givesFor the homogeneous problem, ${\Phi }_{+}/X_{+}$ and ${\Phi }_{-}/X_{-}$ glue to an entire function vanishing at infinity, so only the zero solution exists; this proves uniqueness. Direct verification uses the principal-value integral $-\pi i/t$: the two terms on the left combine to $(t^4+t^3-t^2-t^4+t^3+t^2)/(2t)=t^2$, exactly the right side at $A=1$.