Harmonic Hardy Classes and Fatou Boundary Limits: Examples and Counterexamples

The data. By [L1] the arc $I=I_h({\zeta }_0)$ is open with $m(I)=2h\in (0,1)$, because $0<h<\frac{1}{2}$. Thus $I$ is a nonempty proper open arc, and its complement is a closed arc of measure $1-2h>0$ with nonempty interior. Hence $f={\mathbf{1}}_I$ is a Borel function with $0\le f\le 1$, $f=1$ exactly on $I$ and $f=0$ exactly on $\mathbb{T}\setminus I$, and $\mathrm{\parallel }f{\mathrm{\parallel }}_{L^p}^p={\int }_{\mathbb{T}}fdm=m(I)$ for $1\le p<\mathrm{\infty }$ while $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=1$ and $\mathrm{\parallel }f{\mathrm{\parallel }}_1=m(I)$. The two endpoints of $I$ are ${\zeta }_{\pm }:=\phi ([t_0\pm h])$; they are the points with $d({\zeta }_0,{\zeta }_{\pm })=h$ and do not belong to $I$.

The chord bound. For $\mathrm{\mid }x\mathrm{\mid }\le \frac{1}{2}$ one has $\mathrm{\mid }{e}^{2\pi ix}-1\mathrm{\mid }=2\mathrm{\mid }\sin (\pi x)\mathrm{\mid }\ge 2\mathrm{\mid }x\mathrm{\mid }$: indeed $\mathrm{\mid }{e}^{2\pi ix}-1{\mathrm{\mid }}^2=4{\sin }^2(\pi x)$ by [L6], and $\pi \mathrm{\mid }x\mathrm{\mid }\le \frac{\pi }{2}<2$ with $\sin y\ge y\mathrm{/}3$ for $0<y\le 2$ gives $\mathrm{\mid }\sin (\pi x)\mathrm{\mid }=\sin (\pi \mathrm{\mid }x\mathrm{\mid })\ge \pi \mathrm{\mid }x\mathrm{\mid }\mathrm{/}3\ge \mathrm{\mid }x\mathrm{\mid }$ since $\pi >3$. Consequently, for all $\zeta ,\eta \in \mathbb{T}$ with $d(\zeta ,\eta )\le \frac{1}{2}$, writing $\zeta =\phi ([t])$, $\eta =\phi ([s])$ and choosing $k$ with $\mathrm{\mid }s-t-k\mathrm{\mid }=d(\zeta ,\eta )$ gives $\mathrm{\mid }\phi (\eta )-\phi (\zeta )\mathrm{\mid }=\mathrm{\mid }{e}^{2\pi i(s-t-k)}-1\mathrm{\mid }\ge 2d(\zeta ,\eta )$.

Local concentration of the kernel. Let $\zeta \in \mathbb{T}$ and $0<\delta \le \frac{1}{2}$, and let $z\in \mathbb{D}$ with $\mathrm{\mid }z-\phi (\zeta )\mathrm{\mid }\le \delta$. For every $\eta \in \mathbb{T}$ with $d(\eta ,\zeta )\ge \delta$ step 1.2 gives $\mathrm{\mid }\phi (\eta )-\phi (\zeta )\mathrm{\mid }\ge 2\delta$, hence $\mathrm{\mid }\phi (\eta )-z\mathrm{\mid }\ge \mathrm{\mid }\phi (\eta )-\phi (\zeta )\mathrm{\mid }-\mathrm{\mid }\phi (\zeta )-z\mathrm{\mid }\ge 2\delta -\delta =\delta$, and therefore $P(z,\eta )\le (1-\mathrm{\mid }z{\mathrm{\mid }}^2)\mathrm{/}{\delta }^2$ by [L2]. Thus $\sup \{P(z,\eta ):\eta \in \mathbb{T},d(\eta ,\zeta )\ge \delta \}\le (1-\mathrm{\mid }z{\mathrm{\mid }}^2)\mathrm{/}{\delta }^2\to 0$ as $z\to \phi (\zeta )$ inside $\mathbb{D}$.

The extension and the strict bounds. The Poisson integral $u=P[f]$ is defined, and [L2] gives $u(z)={\int }_{\mathbb{T}}P(z,\eta )f(\eta )dm(\eta )={\int }_{I}P(z,\eta )dm(\eta )$ for every $z\in \mathbb{D}$. Since $P(z,\cdot )>0$ on $\mathbb{T}$ and $m(I)>0$, the integral over $I$ is strictly positive: if it vanished, then $P(z,\cdot ){\mathbf{1}}_I=0$ $m$-almost everywhere by [L7], contradicting positivity on the non-null set $I$. Likewise $P(z,\cdot )>0$ on the non-null set $\mathbb{T}\setminus I$, so ${\int }_{\mathbb{T}\setminus I}P(z,\cdot )dm>0$; additivity of the nonnegative integral over the disjoint union $\mathbb{T}=I\bigsqcup (\mathbb{T}\setminus I)$ and the unit mass ${\int }_{\mathbb{T}}P(z,\eta )dm(\eta )=1$ of [L3] therefore give $u(z)={\int }_{\mathbb{T}}Pdm-{\int }_{\mathbb{T}\setminus I}Pdm<1$. Hence $0<u(z)<1$ for every $z\in \mathbb{D}$, and $u$ is complex harmonic by [L4]; being real-valued, it is a real harmonic function.

The radial limit at an endpoint is one half. Let ${\zeta }_{+}=\phi ([t_0+h])$; for $0\le r<1$ the radial Poisson representation of [L2] and the integral formula of [L1] give $u(r{\zeta }_{+})={\int }_0^1{P}_r(t_0+h-\tau ){\mathbf{1}}_{\{d([\tau ],[t_0])<h\}}d\tau$, and translation invariance of $m$ moves the centre to $[0]$, giving ${\int }_0^1{P}_r(h-s){\mathbf{1}}_{\{d([s],[0])<h\}}ds$. Here the set in $[0,1)$ is $[0,h)\cup (1-h,1)$. Split the integral over those two intervals and on the second put $v=s-1$; periodicity gives $P_r(h-s)=P_r(h-v)$, so that piece equals ${\int }_{-h}^0{P}_r(h-v)dv$. The linear substitutions are licensed by A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions; all integrands are bounded on these finite intervals, and interval endpoints have Lebesgue measure zero. Combining the pieces and then setting $w=h-s$ yields $u(r{\zeta }_{+})={\int }_{-h}^h{P}_r(h-s)ds={\int }_0^{2h}{P}_r(w)dw$. By evenness of $P_r$, ${\int }_0^{2h}{P}_r=\frac{1}{2}{\int }_{-2h}^{2h}{P}_r$. If $2h\le \frac{1}{2}$, then $[-2h,2h]$ lies in a fundamental interval $[-\frac{1}{2},\frac{1}{2}]$, and its complement there stays at torus distance at least $2h>0$ from $0$; [L3] therefore gives ${\int }_{-2h}^{2h}{P}_r\to 1$ (with the complement empty when $2h=\frac{1}{2}$). If $\frac{1}{2}<2h<1$, split $[-2h,2h]$ into $[-\frac{1}{2},\frac{1}{2}]$ and the two extra intervals $[-2h,-\frac{1}{2}]$ and $[\frac{1}{2},2h]$. The fundamental interval has integral $1$, while on the extra intervals the torus distance to $0$ is at least $1-2h>0$, so their integrals tend to $0$ by [L3]. Thus in either case ${\int }_{-2h}^{2h}{P}_r\to 1$ and $u(r{\zeta }_{+})\to \frac{1}{2}$. The same computation at ${\zeta }_{-}$, using evenness, gives $u(r{\zeta }_{-})\to \frac{1}{2}$.

No two-sided limit of the indicator at an endpoint. Fix $m$ so large that $\frac{1}{m}<\min \{h,\frac{1}{2}-h\}$. Then ${\eta }_m:=\phi ([t_0+h-\frac{1}{m}])$ satisfies $d({\zeta }_0,{\eta }_m)=h-\frac{1}{m}<h$ and so lies in $I$, whereas ${\eta }_m^{\mathrm{\prime }}:=\phi ([t_0+h+\frac{1}{m}])$ satisfies $d({\zeta }_0,{\eta }_m^{\mathrm{\prime }})=h+\frac{1}{m}>h$ and so lies in $\mathbb{T}\setminus I$; both sequences converge to ${\zeta }_{+}$ as $m\to \mathrm{\infty }$ by continuity of $\phi$. Hence $f({\eta }_m)=1$ and $f({\eta }_m^{\mathrm{\prime }})=0$ eventually, and $f$ has no limit at ${\zeta }_{+}$; the same argument with $t_0-h$ applies to the other endpoint ${\zeta }_{-}$.

The $L^p$ bound. Since $u_r=P_r\ast f$ by [L2], the contraction inequality of [L4] and step 1.1 give $\mathrm{\parallel }{u}_r{\mathrm{\parallel }}_{L^p(\mathbb{T},m)}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{L^p}=m(I{)}^{1\mathrm{/}p}$ for every $1\le p<\mathrm{\infty }$ and every $0\le r<1$.

The $L^1$ identity. For $0<r<1$ one has ${\int }_{\mathbb{T}}{u}_{r}dm=\frac{1}{2\pi }{\int }_0^{2\pi }u(r{e}^{i\theta })d\theta =u(0)$: the first equality is the torus normalization of [L1], and the second is the circle mean-value property [L5] applied to the real harmonic $u$ on the disc $\overline{D(0,r)}\subseteq \mathbb{D}$. For $r=0$ the identity $u_0\equiv u(0)$ gives the same value. Moreover $u(0)={\int }_{I}P(0,\eta )dm(\eta )={\int }_{I}1dm=m(I)$ because $P(0,\eta )=1$ by [L2]. Since $u_r\ge 0$ by step 2.2, $\mathrm{\parallel }{u}_r{\mathrm{\parallel }}_{L^1(\mathbb{T},m)}={\int }_{\mathbb{T}}{u}_{r}dm=m(I)=2h$ for every $0\le r<1$.

The boundary limits at interior points. Let $\zeta$ be an interior point of $I$: then $d({\zeta }_0,\zeta )<h$, and choosing $0<\delta <\frac{1}{2}$ with $d({\zeta }_0,\zeta )+\delta <h$ gives $d({\zeta }_0,\eta )<h$ for every $\eta$ with $d(\eta ,\zeta )<\delta$ by the triangle inequality for the circular distance; that is, $I_{\delta }(\zeta )\subseteq I$ and $f=1$ there. For $z\in \mathbb{D}$ with $\mathrm{\mid }z-\phi (\zeta )\mathrm{\mid }\le \delta$ one then has $\mathrm{\mid }u(z)-1\mathrm{\mid }=\mid {\int }_{\mathbb{T}}(f(\eta )-1)P(z,\eta )dm(\eta )\mid \le {\int }_{\mathbb{T}\setminus I_{\delta }(\zeta )}P(z,\eta )dm(\eta )\le \sup \{P(z,\eta ):d(\eta ,\zeta )\ge \delta \}$, which tends to $0$ as $z\to \phi (\zeta )$ by step 2.1. Hence $u(z)\to 1$ as $z\to \zeta$ without restriction, and in particular nontangentially. If instead $\zeta$ is an interior point of $\mathbb{T}\setminus I$, choose $\delta$ with $I_{\delta }(\zeta )\cap I=\mathrm{\varnothing }$; then $f=0$ on $I_{\delta }(\zeta )$ and the same estimate without the term $1$ gives $\mathrm{\mid }u(z)\mathrm{\mid }\le \sup \{P(z,\eta ):d(\eta ,\zeta )\ge \delta \}\to 0$, so $u(z)\to 0$.

Assembly. Step 2.2 gives the harmonic extension with $0<u<1$; steps 3.2 and 3.1 give the norm identities $\mathrm{\parallel }{u}_r{\mathrm{\parallel }}_1=m(I)$ and $\mathrm{\parallel }{u}_r{\mathrm{\parallel }}_p\le m(I{)}^{1\mathrm{/}p}$; step 3.3 gives the unrestricted, hence nontangential, boundary values $1$ on interior points of $I$ and $0$ on interior points of the complement; step 2.3 gives the radial limit $\frac{1}{2}$ at each endpoint, while step 2.4 shows that the chosen indicator representative $f={\mathbf{1}}_I$ has no two-sided pointwise limit at either endpoint. ∎

Integrating bounded Borel functions against ${\delta }_{{\zeta }_0}$. Since ${\delta }_{{\zeta }_0}$ is a probability measure, $\mathrm{\mid }{\delta }_{{\zeta }_0}\mathrm{\mid }={\delta }_{{\zeta }_0}$ and ${\delta }_{{\zeta }_0}(\mathbb{T})=1$ by [L6] and [L5]. Let $h$ be a bounded Borel function on $\mathbb{T}$ and let $s:=h({\zeta }_0){\mathbf{1}}_{\mathbb{T}}$, a complex simple function with simple integral ${\int }_{\mathbb{T}}sd{\delta }_{{\zeta }_0}=h({\zeta }_0){\delta }_{{\zeta }_0}(\mathbb{T})=h({\zeta }_0)$. The constant sequence $(s)$ is admissible in the definition of ${\int }_{\mathbb{T}}hd{\delta }_{{\zeta }_0}$, because $\mathrm{\mid }h-s\mathrm{\mid }=\mathrm{\mid }h-h({\zeta }_0)\mathrm{\mid }$ is a nonnegative measurable function with ${\int }_{\mathbb{T}}\mathrm{\mid }h-s\mathrm{\mid }d\mathrm{\mid }{\delta }_{{\zeta }_0}\mathrm{\mid }=0$: the integrand vanishes at ${\zeta }_0$ and is supported on $\mathbb{T}\setminus \{{\zeta }_0\}$, which is ${\delta }_{{\zeta }_0}$-null by [L5], so its integral vanishes by A nonnegative integral over a null set vanishes. Hence ${\int }_{\mathbb{T}}hd{\delta }_{{\zeta }_0}=h({\zeta }_0)$.

The function $u$ is the translate of the kernel. Fix $z\in \mathbb{D}$. The function $\zeta \mapsto P(z,\zeta )$ is continuous on $\mathbb{T}$ by [L2], hence bounded Borel, so step 1.1 and the definition of the Poisson integral give $u(z)={\int }_{\mathbb{T}}P(z,\zeta )d{\delta }_{{\zeta }_0}(\zeta )=P(z,{\zeta }_0)=(1-\mathrm{\mid }z{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }\phi ({\zeta }_0)-z{\mathrm{\mid }}^2$, and this is strictly positive because $\mathrm{\mid }\phi ({\zeta }_0)-z\mathrm{\mid }\ge 1-\mathrm{\mid }z\mathrm{\mid }>0$.

Membership and norm. By [L7] the Dirac measure is a finite regular complex Borel measure, so the converse clause of [L3] shows that $u=P[{\delta }_{{\zeta }_0}]$ lies in $h^1(\mathbb{D})$ with $\mathrm{\parallel }u{\mathrm{\parallel }}_{h^1}=\mathrm{\mid }{\delta }_{{\zeta }_0}\mathrm{\mid }(\mathbb{T})={\delta }_{{\zeta }_0}(\mathbb{T})=1$, and $u$ is harmonic by [L4].

Limits at the other boundary points. Let $\zeta \in \mathbb{T}\setminus \{{\zeta }_0\}$ and let $z_m\to \zeta$ with $z_m\in \mathbb{D}$. Then step 2.1 gives $u(z_m)=(1-\mathrm{\mid }{z}_m{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }\phi ({\zeta }_0)-z_m{\mathrm{\mid }}^2$; here $1-\mathrm{\mid }{z}_m{\mathrm{\mid }}^2\to 0$, while by injectivity of $\phi$ and $\zeta \mathrm{\ne }{\zeta }_0$ one has $\phi ({\zeta }_0)\mathrm{\ne }\zeta$, so $\mathrm{\mid }\phi ({\zeta }_0)-z_m{\mathrm{\mid }}^2\to \mathrm{\mid }\phi ({\zeta }_0)-\zeta {\mathrm{\mid }}^2>0$. Hence $u(z_m)\to 0$: the limit is $0$ at every boundary point other than ${\zeta }_0$, along arbitrary sequences inside the disc.

The radial blow-up at ${\zeta }_0$. For $0\le r<1$ one has $\phi ({\zeta }_0)-r\phi ({\zeta }_0)=(1-r)\phi ({\zeta }_0)$ with $\mathrm{\mid }\phi ({\zeta }_0)\mathrm{\mid }=1$, so step 2.1 gives $u(r{\zeta }_0)=\frac{1-r^2}{(1-r{)}^2}=\frac{1+r}{1-r}$. As $r\uparrow 1$ the numerator tends to $2$ and the denominator to $0$ through positive values, so $u(r{\zeta }_0)\to +\mathrm{\infty }$; in particular $u$ is unbounded on $\mathbb{D}$.

No $L^1$ density. Suppose $f\in L^1(\mathbb{T},m)$ satisfied $P[f]=u$. Then $P[fm]=P[f]=u=P[{\delta }_{{\zeta }_0}]$, and by [L7] the density measure $fm$ is a finite complex measure with $\mathrm{\mid }fm\mathrm{\mid }(\mathbb{T})=\mathrm{\parallel }f{\mathrm{\parallel }}_1<\mathrm{\infty }$, hence a finite regular complex Borel measure; being finite and regular it is one of the measures to which the uniqueness clause of [L3] applies, so $fm={\delta }_{{\zeta }_0}$. Evaluating both sides at the singleton $\{{\zeta }_0\}$ gives $0={\int }_{\{{\zeta }_0\}}fdm=(fm)(\{{\zeta }_0\})={\delta }_{{\zeta }_0}(\{{\zeta }_0\})=1$: the middle integral vanishes because $m(\{{\zeta }_0\})=0$ and $f\in L^1(\mathbb{T},m)$, while the last value is $1$ by [L5]. This contradiction shows that no $f\in L^1(\mathbb{T},m)$ represents $u$.

Assembly. Steps 2.1, 3.1, 4.1, 3.2 and 3.3 establish, respectively, the identification $u(z)=P(z,{\zeta }_0)>0$, the membership $u\in h^1(\mathbb{D})$ with norm $1$ and harmonicity, the absence of a representing $L^1$ density, the boundary limit $0$ at every point other than ${\zeta }_0$, and the radial formula $u(r{\zeta }_0)=(1+r)\mathrm{/}(1-r)\to +\mathrm{\infty }$. So the boundary atom ${\delta }_{{\zeta }_0}$ produces an unbounded positive $h^1$ function whose boundary measure is singular with respect to $m$ and which therefore has no $L^1$ density; the Axiom of Choice is used only through the representation theorem [L3] and the countable-choice regularity corollary [L7]. ∎

Geometry of the arcs. Put $t_n=2^{-n}$ and $w_n=2^{-3n}$ for $n\ge 3$; then $0<t_n-w_n$, $t_n+w_n\le t_3+w_3=\frac{65}{512}<\frac{1}{4}$, and $w_n\le \frac{1}{64}{t}_n$. The arcs $I_n=I_{w_n}(q(t_n))$ are pairwise disjoint: for $m>n$ one has $\mathrm{\mid }{t}_n-t_m\mathrm{\mid }\ge t_{n+1}$ while $w_m+w_n\le w_{n+1}+w_n=\frac{9}{8}{w}_n$, and $t_{n+1}=2^{-n-1}\ge \frac{9}{8}\cdot 2^{-3n}=\frac{9}{8}{w}_n$; all arcs lie in $\{d(q(0),\cdot )<\frac{1}{4}\}$, so the circular distance between $q(t_n)$ and $q(t_m)$ is the number $\mathrm{\mid }{t}_n-t_m\mathrm{\mid }$. Consequently $m(I_n)=2w_n$, the sets $I_n$ are pairwise disjoint, and $E\subseteq I_{1\mathrm{/}4}(q(0))$. Moreover $q(t_n)\in I_n$, while $q(\frac{3}{2}{t}_n)\mathrm{\notin }{I}_m$ for every $m\ge 3$: for $m=n$ the circular distance is $\frac{1}{2}{t}_n>w_n$; for $m>n$ it is at least the distance to $t_{n+1}$, namely $t_n>w_m$; and for $m<n$, $t_m\ge 2t_n$, so the distance is $t_m-\frac{3}{2}{t}_n\ge \frac{1}{4}{t}_m>w_m$, since $w_m=t_m^3\le \frac{1}{64}{t}_m$.