Poisson nontangential maximal function is controlled by circle maximal averages

Let $z\in {\mathrm{\Gamma }}_A(\zeta )$, $r=\mathrm{\mid }z\mathrm{\mid }$, and $\eta \in \mathbb{T}$. Put $u:=\mathrm{Re}(\bar{\zeta }z)\le r$. Since $\mathrm{\mid }r\zeta -z{\mathrm{\mid }}^2=2r^2-2ru$ and $\mathrm{\mid }\zeta -z{\mathrm{\mid }}^2=1+r^2-2u$, the identity $2r^2-2ru\le 1+r^2-2u$ is equivalent to $(1-r)(1+r-2u)\ge 0$, which holds because $1-r>0$ and $2u\le 2r\le 1+r$; hence $\mathrm{\mid }r\zeta -z\mathrm{\mid }\le \mathrm{\mid }\zeta -z\mathrm{\mid }$. Therefore $\mathrm{\mid }r\zeta -\eta \mathrm{\mid }\le \mathrm{\mid }r\zeta -z\mathrm{\mid }+\mathrm{\mid }z-\eta \mathrm{\mid }\le \mathrm{\mid }\zeta -z\mathrm{\mid }+\mathrm{\mid }z-\eta \mathrm{\mid }\le A(1-r)+\mathrm{\mid }z-\eta \mathrm{\mid }\le (A+1)\mathrm{\mid }z-\eta \mathrm{\mid },$ where the cone condition gives $\mathrm{\mid }\zeta -z\mathrm{\mid }<A(1-r)$ and the reverse triangle inequality gives $\mathrm{\mid }z-\eta \mathrm{\mid }\ge 1-\mathrm{\mid }z\mathrm{\mid }=1-r$. The case $z=0$ is included: then $r=0$, $u=0$ and $\mathrm{\mid }r\zeta -\eta \mathrm{\mid }=1\le A+1=(A+1)\mathrm{\mid }z-\eta \mathrm{\mid }$.

Fix $z=r\zeta$ with $0<r<1$ and put $g(\delta ):=(1-r^2)\mathrm{/}(1-2r\cos (2\pi \delta )+r^2)$ for $\delta \in [0,\frac{1}{2}]$. By [L2], $P(z,\eta )=g(d(\zeta ,\eta ))$ for every $\eta \in \mathbb{T}$, and by [L3] the function $g$ is continuous and strictly decreasing on $[0,\frac{1}{2}]$. Given $\epsilon >0$, choose $0={\delta }_0<{\delta }_1<\cdots <{\delta }_m=\frac{1}{2}$ with $g({\delta }_{j-1})-g({\delta }_j)\le \epsilon$ for all $j$ (continuity on a compact interval); put $c_0:=g(\frac{1}{2})$ and $c_j:=g({\delta }_{j-1})-g({\delta }_j)\ge 0$. Then for every $\delta \in [0,\frac{1}{2}]$ one has $g(\delta )\le \varphi (\delta ):=c_0+{\sum }_{j=1}^m{c}_j{\mathbf{1}}_{[0,{\delta }_j)}(\delta )\le g(\delta )+\epsilon ,$ because on $[{\delta }_{k-1},{\delta }_k)$ the function $\varphi$ equals $g({\delta }_{k-1})$ while $g({\delta }_k)\le g(\delta )\le g({\delta }_{k-1})$ and $g({\delta }_{k-1})-g({\delta }_k)=c_k\le \epsilon$; on the single point $\delta =\frac{1}{2}$ both sides equal $g(\frac{1}{2})$. Consequently $P(z,\eta )\le c_0+{\sum }_j{c}_j{\mathbf{1}}_{\{d(\zeta ,\eta )<{\delta }_j\}}(\eta )$ for every $\eta$, while integrating the two-sided bound against $m$ and using $m(\mathbb{T})=1$ together with ${\int }_{\mathbb{T}}g(d(\zeta ,\eta ))dm(\eta )={\int }_{\mathbb{T}}P(z,\eta )dm(\eta )=1$ gives $c_0+{\sum }_{j=1}^m{c}_{j}m(\{d(\zeta ,\cdot )<{\delta }_j\})\le 1+\epsilon \mathrm{.}$

By definition of $M_{\mathbb{T}}\mu$ as a supremum over $h\in (0,\frac{1}{2}]$, every arc satisfies $\mathrm{\mid }\mu \mathrm{\mid }(I_h(\zeta ))\le M_{\mathbb{T}}\mu (\zeta )m(I_h(\zeta ))=2h{M}_{\mathbb{T}}\mu (\zeta )$; in particular $\mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})=\mathrm{\mid }\mu \mathrm{\mid }(I_{1\mathrm{/}2}(\zeta ))\le M_{\mathbb{T}}\mu (\zeta )$. Moreover each set $\{d(\zeta ,\cdot )<\delta \}$ with $\delta \in (0,\frac{1}{2}]$ satisfies $m(\{d<\delta \})=2\delta$ and $\mathrm{\mid }\mu \mathrm{\mid }(\{d<\delta \})\le M_{\mathbb{T}}\mu (\zeta )m(\{d<\delta \})$: for $\delta <\frac{1}{2}$ the set is the arc $I_{\delta }(\zeta )$, and for $\delta =\frac{1}{2}$ it is $\mathbb{T}$ minus the antipode, whose $m$-measure is $1$ and whose $\mathrm{\mid }\mu \mathrm{\mid }$-measure is at most $\mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})\le M_{\mathbb{T}}\mu (\zeta )$.

Put $M:=M_{\mathbb{T}}\mu (\zeta )$. If $M=+\mathrm{\infty }$, the desired bound is automatic. Assume $M<+\mathrm{\infty }$. For $0<r<1$ the pointwise bound of step 1.2 and the estimates of step 1.3 give $\mathrm{\mid }P[\mu ](r\zeta )\mathrm{\mid }\le {\int }_{\mathbb{T}}P(r\zeta ,\eta )d\mathrm{\mid }\mu \mathrm{\mid }(\eta )\le c_0\mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})+{\sum }_{j=1}^m{c}_j\mathrm{\mid }\mu \mathrm{\mid }(\{d(\zeta ,\cdot )<{\delta }_j\})\le M(c_0+{\sum }_{j=1}^m{c}_{j}m(\{d(\zeta ,\cdot )<{\delta }_j\}\left)\right)\le (1+\epsilon )M,$ where [L4] supplies the first inequality. Since $\epsilon >0$ was arbitrary, ${\int }_{\mathbb{T}}P(r\zeta ,\eta )d\mathrm{\mid }\mu \mathrm{\mid }(\eta )\le M$ and hence $\mathrm{\mid }P[\mu ](r\zeta )\mathrm{\mid }\le M$. For $r=0$, $P(0,\eta )=1$ and step 1.3 give ${\int }_{\mathbb{T}}P(0,\eta )d\mathrm{\mid }\mu \mathrm{\mid }(\eta )=\mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})\le M$, while [L4] gives $\mathrm{\mid }P[\mu ](0)\mathrm{\mid }=\mathrm{\mid }\mu (\mathbb{T})\mathrm{\mid }\le \mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})$.

For $z\in {\mathrm{\Gamma }}_A(\zeta )$ with $\mathrm{\mid }z\mathrm{\mid }=r$ and every $\eta \in \mathbb{T}$, step 1.1 gives $\mathrm{\mid }\eta -r\zeta \mathrm{\mid }\le (A+1)\mathrm{\mid }\eta -z\mathrm{\mid }$, hence $P(z,\eta )=\frac{1-r^2}{\mathrm{\mid }\eta -z{\mathrm{\mid }}^2}\le (A+1{)}^2\frac{1-r^2}{\mathrm{\mid }\eta -r\zeta {\mathrm{\mid }}^2}=(A+1{)}^{2}P(r\zeta ,\eta )\mathrm{.}$

Combining steps 2.1 and 2.2, for every $z\in {\mathrm{\Gamma }}_A(\zeta )$ with $\mathrm{\mid }z\mathrm{\mid }=r$, $\mathrm{\mid }P[\mu ](z)\mathrm{\mid }\le {\int }_{\mathbb{T}}P(z,\eta )d\mathrm{\mid }\mu \mathrm{\mid }(\eta )\le (A+1{)}^2{\int }_{\mathbb{T}}P(r\zeta ,\eta )d\mathrm{\mid }\mu \mathrm{\mid }(\eta )\le (A+1{)}^2{M}_{\mathbb{T}}\mu (\zeta ),$ the first inequality by [L4]. Taking the supremum over $z\in {\mathrm{\Gamma }}_A(\zeta )$ gives $N_A(P[\mu ])(\zeta )\le (A+1{)}^2{M}_{\mathbb{T}}\mu (\zeta )$. For $f\in L^1(\mathbb{T},m)$ the identities $P[f]=P[fm]$ and $M_{\mathbb{T}}f=M_{\mathbb{T}}(fm)$ of [L1] give $N_A(P[f])(\zeta )\le (A+1{)}^2{M}_{\mathbb{T}}f(\zeta )$, completing the proof.