The fixed-support Cauchy transform and its Hölder bounds

For $z\mathrm{\ne }0$, direct differentiation of $\mathrm{\Gamma }$ gives $a(z):={\mathrm{\partial }}_z\mathrm{\Gamma }(z)=-1\mathrm{/}(4\pi z)$; polar coordinates give ${\int }_{D_r}\mathrm{\mid }a\mathrm{\mid }dA=r\mathrm{/}2$. Integrating by parts outside $D_{\epsilon }$ against a compactly supported smooth test function leaves an inner boundary term bounded by $C\epsilon \mathrm{\mid }\log \epsilon \mathrm{\mid }$, which tends to zero, so the distributional derivative of $\mathrm{\Gamma }$ is the regular distribution of $a$.

Since $Nq\in C^2$ and $-\mathrm{\Delta }Nq=q$, the identity $\mathrm{\Delta }=4{\mathrm{\partial }}_z{\mathrm{\partial }}_{\bar{z}}$ gives ${\mathrm{\partial }}_{\bar{z}}Tq=-4{\mathrm{\partial }}_{\bar{z}}{\mathrm{\partial }}_{z}Nq=q$ pointwise. On every compact subset of $\mathbb{C}\setminus {\bar{D}}_2$, the Newtonian kernel $\mathrm{\Gamma }(z-\zeta )$ and all its $z$-derivatives are bounded uniformly for $\zeta \in {\bar{D}}_2$; differentiation under the integral sign in $Nq$ therefore makes $Nq$, and hence $Tq$, smooth there.

By Fubini and integration by parts in the compactly supported $q$ variable, for every multi-index $\beta$ with $\mathrm{\mid }\beta \mathrm{\mid }\le k$ the distributional identity $D^{\beta }Nq=N(D^{\beta }q)$ holds. The right side is $C_{\mathrm{l}\mathrm{o}\mathrm{c}}^{2,\alpha }$ by [F4]. Starting with $Nq\in C^2$, induction on $\mathrm{\mid }\beta \mathrm{\mid }$ identifies each already-classical derivative $D^{\beta }Nq$ with this continuous representative: distributional injectivity gives equality almost everywhere, and [F8] rules out a nonzero continuous difference on any ball. Each such derivative is then $C^2$. The local estimate in [F4], applied to each $D^{\beta }q$ with support in ${\bar{D}}_2$, yields $Nq\in C_{\mathrm{l}\mathrm{o}\mathrm{c}}^{k+2,\alpha }$ with its norm on ${\bar{D}}_R$ bounded by $C_{k,\alpha ,R}\mathrm{\parallel }q{\mathrm{\parallel }}_{C^{k,\alpha }}$. Since $Tq=-4{\mathrm{\partial }}_{z}Nq$, this gives the asserted $C^{k+1,\alpha }(D_R)$ bound.

The cancellation formula [F5], combined as $Sq=-({\mathrm{\partial }}_{xx}-2i{\mathrm{\partial }}_{xy}-{\mathrm{\partial }}_{yy})Nq$, cancels the two diagonal correction terms. Away from zero the resulting kernel is $-4{\mathrm{\partial }}_z^2\mathrm{\Gamma }(w)=-1\mathrm{/}(\pi w^2)$, so $Sq(z)=-(1\mathrm{/}\pi )\mathrm{p.v.}\int q(\zeta )\mathrm{/}(z-\zeta {)}^{2}dA(\zeta )$. The integral of $(z-\zeta {)}^{-2}$ over every centered annulus is zero because its angular factor is $e^{-2i\theta }$; hence subtracting $q(z)$ only on $\mathrm{\mid }z-\zeta \mathrm{\mid }<1$ gives the displayed subtracted formula. Its near integral is bounded absolutely by $C[q{]}_{0,\alpha }{\int }_0^1{r}^{\alpha -1}dr$, and the far integral is absolutely finite because it avoids the singularity and $q$ has compact support.

For every compactly supported smooth test function $\phi$, Fubini and the distributional derivative identity in step 1.1 give $⟨{\mathrm{\partial }}_{z}Nq,\phi ⟩=⟨a\ast q,\phi ⟩$, with $(a\ast q)(z)={\int }_{\mathbb{C}}a(z-\zeta )q(\zeta )dA(\zeta )\mathrm{.}$ This integral is absolutely finite for each $z$, since $q$ is bounded, supported in ${\bar{D}}_2$, and $a$ is locally integrable. It is continuous: on a compact set of $z$-values let $\delta =\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }<1$; the two disks of radius $2\delta$ around $z,z^{\mathrm{\prime }}$ contribute at most $C\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}\delta$, while on their complement $\mathrm{\mid }\mathrm{\nabla }a(w)\mathrm{\mid }\le C\mathrm{\mid }w{\mathrm{\mid }}^{-2}$ and the mean value theorem bounds the difference by $C\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}\delta \log (C_0\mathrm{/}\delta )$ for a fixed $C_0$. Since $Nq\in C^2$, both sides are continuous; [F6] makes them equal almost everywhere, and [F8] then makes them equal everywhere. Therefore $Tq=-4{\mathrm{\partial }}_{z}Nq=(1\mathrm{/}\pi )\int q(\zeta )\mathrm{/}(z-\zeta )dA(\zeta )$.

For the global Hölder seminorm when $k=0$, put $K(w)=-1\mathrm{/}(\pi w^2)$ and $k_{ij}(w)={\mathrm{\partial }}_i{\mathrm{\partial }}_j\mathrm{\Gamma }(w)$. Let $\mathrm{\Omega }$ be a disk with $\mathrm{supp}q⋐\mathrm{\Omega }$ and $x\in \mathrm{\Omega }$, and choose a larger disk $D_s(x)\supset \bar{\mathrm{\Omega }}$. On the outer circle the explicit derivative ${\mathrm{\partial }}_i\mathrm{\Gamma }(w)=-w_i\mathrm{/}(2\pi \mathrm{\mid }w{\mathrm{\mid }}^2)$ and polar symmetry give ${\int }_{\mathrm{\partial }{D}_s(x)}{\mathrm{\partial }}_i\mathrm{\Gamma }(x-y){\nu }_j(y)dS(y)={\delta }_{ij}\mathrm{/}2$. Also $k_{ij}(x-y)=-{\mathrm{\partial }}_{y_j}{\mathrm{\partial }}_i\mathrm{\Gamma }(x-y)$, so the divergence theorem gives ${\int }_{D_s(x)\setminus \mathrm{\Omega }}{k}_{ij}(x-y)dA(y)=-{\delta }_{ij}\mathrm{/}2+g_{ij,\mathrm{\Omega }}(x)$, where $g_{ij,\mathrm{\Omega }}(x):={\int }_{\mathrm{\partial }\mathrm{\Omega }}{\mathrm{\partial }}_i\mathrm{\Gamma }(x-y){\nu }_j(y)dS(y)\mathrm{.}$ Split the centered-disk cancellation formula [F5] into $\mathrm{\Omega }$ and $D_s(x)\setminus \mathrm{\Omega }$; on the latter $q(y)=0$, so the ${\delta }_{ij}\mathrm{/}2$ terms cancel and ${\mathrm{\partial }}_i{\mathrm{\partial }}_{j}Nq(x)={\int }_{\mathrm{\Omega }}{k}_{ij}(x-y)(q(y)-q(x))dA(y)-q(x)g_{ij,\mathrm{\Omega }}(x)\mathrm{.}$ Taking the linear combination $-{\mathrm{\partial }}_{xx}+2i{\mathrm{\partial }}_{xy}+{\mathrm{\partial }}_{yy}$ gives the corresponding formula for $Sq$ with kernel $K$ and boundary factor $G_{\mathrm{\Omega }}=-g_{xx,\mathrm{\Omega }}+2i{g}_{xy,\mathrm{\Omega }}+g_{yy,\mathrm{\Omega }}$.

Fix distinct $x,x^{\mathrm{\prime }}$, put $\delta =\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }$ and $m=(x+x^{\mathrm{\prime }})\mathrm{/}2$, and take $\mathrm{\Omega }=D_R(m)$ with $\mathrm{supp}q⋐\mathrm{\Omega }$ and $R\ge 2\delta$. Reflection through $m$ sends $x$ to $x^{\mathrm{\prime }}$, reverses both ${\mathrm{\partial }}_i\mathrm{\Gamma }(x-y)$ and the normal ${\nu }_j(y)$, and preserves arc length, so $G_{\mathrm{\Omega }}(x)=G_{\mathrm{\Omega }}(x^{\mathrm{\prime }})$. Also $\mathrm{\mid }{G}_{\mathrm{\Omega }}(x)\mathrm{\mid }\le C$, since $\mathrm{\mid }x-y\mathrm{\mid }\ge 3R\mathrm{/}4$ on $\mathrm{\partial }\mathrm{\Omega }$, $\mathrm{\mid }\mathrm{\nabla }\mathrm{\Gamma }(w)\mathrm{\mid }\le C\mathrm{/}\mathrm{\mid }w\mathrm{\mid }$, and $\mathrm{\partial }\mathrm{\Omega }$ has length $2\pi R$. Thus the boundary-term difference is at most $C[q{]}_{0,\alpha }{\delta }^{\alpha }$.

Split the integral difference over $D_{\delta }(m)$ and $\mathrm{\Omega }\setminus {\bar{D}}_{\delta }(m)$. On the inner disk, $\mathrm{\mid }K(x-y)(q(y)-q(x))\mathrm{\mid }\le C[q{]}_{0,\alpha }\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha -2}$ and likewise for $x^{\mathrm{\prime }}$, so polar integration bounds both contributions by $C_{\alpha }[q{]}_{0,\alpha }{\delta }^{\alpha }$. On the outer region, write the difference integrand as $[K(x-y)-K(x^{\mathrm{\prime }}-y)](q(y)-q(x))-K(x^{\mathrm{\prime }}-y)(q(x)-q(x^{\mathrm{\prime }}))\mathrm{.}$ With $\rho =\mathrm{\mid }y-m\mathrm{\mid }\ge \delta$, the segment between $x-y$ and $x^{\mathrm{\prime }}-y$ stays at distance at least $\rho \mathrm{/}2$ from zero; $\mathrm{\mid }\mathrm{\nabla }K(w)\mathrm{\mid }\le C\mathrm{\mid }w{\mathrm{\mid }}^{-3}$ and the mean value theorem bound the first term by $C[q{]}_{0,\alpha }\delta {\rho }^{\alpha -3}$. Its area integral is at most $C[q{]}_{0,\alpha }\delta {\int }_{\delta }^R{\rho }^{\alpha -2}d\rho \le C_{\alpha }[q{]}_{0,\alpha }{\delta }^{\alpha }$, using $\alpha <1$. For the second term, $\mathrm{\mid }q(x)-q(x^{\mathrm{\prime }})\mathrm{\mid }\le [q{]}_{0,\alpha }{\delta }^{\alpha }$ and each real component ${\int }_{\mathrm{\Omega }\setminus {\bar{D}}_{\delta }(m)}{k}_{ij}(x^{\mathrm{\prime }}-y)dA(y)$ is bounded by the divergence theorem: its boundary fluxes are ${\mathrm{\partial }}_i\mathrm{\Gamma }$ on the outer circle and inner circle, each bounded by $C$ using $\mathrm{\mid }{x}^{\mathrm{\prime }}-y\mathrm{\mid }\ge 3R\mathrm{/}4$ on the outer circle and $\mathrm{\mid }{x}^{\mathrm{\prime }}-y\mathrm{\mid }\ge \delta \mathrm{/}2$ on the inner one. This proves $[Sq{]}_{0,\alpha ;{\mathbb{R}}^2}\le C_{\alpha }[q{]}_{0,\alpha ;{\mathbb{R}}^2}$.

On ${\bar{D}}_4$, the local Hessian estimate [F4] bounds $\mathrm{\mid }Sq\mathrm{\mid }$ by $C_{\alpha }\mathrm{\parallel }q{\mathrm{\parallel }}_{C^{0,\alpha }}$. For $\mathrm{\mid }z\mathrm{\mid }\ge 3$, the integral formula gives $\mathrm{\mid }Sq(z)\mathrm{\mid }\le \frac{\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}}{\pi }{\int }_{{\bar{D}}_2}\mathrm{\mid }z-\zeta {\mathrm{\mid }}^{-2}dA(\zeta )\le 4\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }},$ since $\mathrm{\mid }z-\zeta \mathrm{\mid }\ge 1$. Hence $\mathrm{\parallel }Sq{\mathrm{\parallel }}_{C^{0,\alpha }({\mathbb{R}}^2)}\le C_{\alpha }\mathrm{\parallel }q{\mathrm{\parallel }}_{C^{0,\alpha }({\mathbb{R}}^2)}$.

For $k\ge 1$, the regularity in step 1.3 makes $D^{\beta }Sq=S(D^{\beta }q)$ classically for every $\mathrm{\mid }\beta \mathrm{\mid }\le k$: expand $S$ as a linear combination of second derivatives of $Nq$ and commute continuous mixed derivatives using [F6]. Each $D^{\beta }q$ is supported in ${\bar{D}}_2$ and has $C^{0,\alpha }$ norm at most $C_k\mathrm{\parallel }q{\mathrm{\parallel }}_{C^{k,\alpha }}$: at top order this is part of the norm; below top order, the mean-value bound controls pairs at distance at most1 by the next derivatives, while twice the supremum controls pairs farther apart. Applying the seminorm and supremum bounds of steps 3.2 and 4.1 to these finitely many derivatives proves $Sq\in C^{k,\alpha }({\mathbb{R}}^2)$ and the stated constant $M_{k,\alpha }$. The zero datum is included, and all estimates use the strict range $0<\alpha <1$; no endpoint $\alpha =1$ or global $L^p$ bound is claimed.