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A radial Poisson limit does not control a tangential path

Statement refuted

Let $u$ be the Poisson integral of a function on the torus. The implication "if $u (r ζ_{0}) \to L$ as $r ↑ 1$ , then $u (z) \to L$ along every sequence $z \to ζ_{0}$ in $D$ " is false, even for $L = 0$ and even for bounded data with values in ${0, 1}$ . Explicitly, put $t_{n} : = 2^{- n}$ and $w_{n} : = 2^{- 3 n}$ for integers $n \geq 3$ , let $I_{n} : = I_{w_{n}} (q (t_{n}))$ be the centred torus arc of radius $w_{n}$ , set $E : = ⋃_{n \geq 3} I_{n}$ , $f : = 1_{E}$ , and let $u : = P [f]$ be the Poisson integral of the density measure $f m$ . Then $f$ is a bounded Borel function with values in ${0, 1}$ , the boundary values of $f$ have no limit at $φ ([0]) = 1$ , and

$u (r φ ([0])) \to 0$ as $r ↑ 1$ , that is, the radial limit of $u$ at the boundary point $1$ exists and equals $0$ ; while
for $z_{n} : = (1 - w_{n}) φ ([t_{n}])$ one has $u (z_{n}) \geq (π + 1)^{- 2}$ for every $n \geq 3$ , $z_{n} \to 1$ , and $∣ z_{n} - 1 ∣ / (1 - ∣ z_{n} ∣) \to \infty$ .

Thus convergence along the radius to a boundary point does not force convergence along other sequences tending to that point. The failure occurs outside every nontangential region: each $z_{n}$ eventually lies outside every $Γ_{A} (1)$ , so the almost-everywhere nontangential Fatou theorem is not affected.

Facts & Assumptions

Given: Countable choice, the torus data $t_{n}, w_{n}, I_{n}, E, f, u, z_{n}$ of the Statement, and the following facts.

[L1]

The torus $T = R / Z$ is identified with the unit circle by $φ ([t]) = e^{2 π i t}$ , the quotient map $q : R \to T$ is continuous, and $m$ is the normalized Haar probability measure; for $ζ \in T$ and $0 < h < \frac{1}{2}$ , the centred arc is $I_{h} (ζ) = {η \in T : d (ζ, η) < h}$ , while $I_{1 / 2} (ζ) = T$ ; every $I_{h} (ζ)$ with $0 < h \leq \frac{1}{2}$ is open and has $m (I_{h} (ζ)) = 2 h$ ; for $A > 1$ the nontangential region is $Γ_{A} (ζ) = {z \in D : ∣ z - ζ ∣ < A (1 - ∣ z ∣)}$ (The one-dimensional torus and its normalized Haar integral, The circle maximal function and nontangential approach regions).

[L2]

For $f \in L^{1} (T, m)$ the Poisson integral is $P [f] = P [f m]$ , given by $P [f] (z) = \int_{T} P (z, η) f (η) d m (η)$ with $P (z, η) = (1 - ∣ z ∣^{2}) / ∣ φ (η) - z ∣^{2} > 0$ for $z \in D$ and $η \in T$ ; writing $z = r φ ([t])$ and $η = φ ([s])$ gives $P (z, η) = (1 - r^{2}) / (1 - 2 r \cos (2 π (s - t)) + r^{2})$ , and the radial function is $(P_{r} * f) (ζ) = P [f] (r ζ) = \int_{T} P_{r} (ζ - η) f (η) d m (η)$ , where $P_{r} (θ) = (1 - r^{2}) / (1 - 2 r \cos (2 π θ) + r^{2})$ in torus coordinates (The Poisson integral of a finite complex boundary measure, The Poisson kernel on the unit disc, The Poisson kernel is positive, has total mass one, and concentrates at a boundary point).

[L3]

If $f \in L^{\infty} (T, m)$ then $f \in L^{1} (T, m)$ , the Poisson integral $P [f]$ is complex harmonic on $D$ , and $∥ (P [f])_{r} ∥_{\infty} \leq ∥ f ∥_{\infty}$ for every $0 \leq r < 1$ (Poisson extension is an Lp contraction and converges in finite Lp, Complex Holder, Minkowski, and the quotient norm, Complex Lp classes and Euclidean test-function conventions).

[L4]

For $0 < x \leq 2$ one has $\sin x \geq x / 3 > 0$ ; for all real $u, v$ one has $∣ \sin u - \sin v ∣ \leq ∣ u - v ∣$ and $∣ \cos u - \cos v ∣ \leq ∣ u - v ∣$ ; for all real $x$ one has $\sin (- x) = - \sin x$ , $\cos (- x) = \cos x$ and $\sin^{2} x + \cos^{2} x = 1$ ; and $\cos (2 x) = 1 - 2 \sin^{2} x$ while $e^{i x} = \cos x + i \sin x$ with $∣ e^{i x} ∣ = 1$ (Sine is positive and cosine is strictly decreasing on (0,2), with cos 2 at most -1/3, Sine and cosine are $1$ -Lipschitz on $R$ , Parity and the Pythagorean identity for sine and cosine, Double-angle and quadratic power-reduction identities, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[L5]

For a measure space, a nonnegative measurable function $g$ and pairwise disjoint measurable sets $E_{n}$ with union $E$ , monotone convergence gives $\int_{E} g = \sum_{n} \int_{E_{n}} g$ ; the integral of a nonnegative simple function is additive in its canonical decomposition, so $\int_{X} c 1_{F} d μ = c μ (F)$ for $c \geq 0$ ; and if $0 \leq g_{1} \leq g_{2}$ then $\int g_{1} \leq \int g_{2}$ (Integral over a measurable subset, The integral of a nonnegative simple function, Monotone convergence for the integral, Monotonicity and nonnegative homogeneity of the nonnegative integral).

[L6]

If $g \in L^{1} (T, m)$ , then for $m$ -almost every $ζ \in T$ the Poisson integral $P [g]$ converges to $g (ζ)$ along every fixed nontangential region $Γ_{A} (ζ)$ ; the theorem asserts nothing about paths that leave every $Γ_{A} (ζ)$ (Fatou limits for Poisson extensions of L1 boundary data).

Counterexample

technique · direct

1.1givenL1algebra

Geometry of the arcs. Put $t_{n} = 2^{- n}$ and $w_{n} = 2^{- 3 n}$ for $n \geq 3$ ; then $0 < t_{n} - w_{n}$ , $t_{n} + w_{n} \leq t_{3} + w_{3} = \frac{65}{512} < \frac{1}{4}$ , and $w_{n} \leq \frac{1}{64} t_{n}$ . The arcs $I_{n} = I_{w_{n}} (q (t_{n}))$ are pairwise disjoint: for $m > n$ one has $∣ t_{n} - t_{m} ∣ \geq t_{n + 1}$ while $w_{m} + w_{n} \leq w_{n + 1} + w_{n} = \frac{9}{8} w_{n}$ , and $t_{n + 1} = 2^{- n - 1} \geq \frac{9}{8} \cdot 2^{- 3 n} = \frac{9}{8} w_{n}$ ; 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 $∣ t_{n} - t_{m} ∣$ . Consequently $m (I_{n}) = 2 w_{n}$ , the sets $I_{n}$ are pairwise disjoint, and $E \subseteq I_{1 / 4} (q (0))$ . Moreover $q (t_{n}) \in I_{n}$ , while $q (\frac{3}{2} t_{n}) \notin I_{m}$ for every $m \geq 3$ : 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 $m < n$ , $t_{m} \geq 2 t_{n}$ , so the distance is $t_{m} - \frac{3}{2} t_{n} \geq \frac{1}{4} t_{m} > w_{m}$ , since $w_{m} = t_{m}^{3} \leq \frac{1}{64} t_{m}$ .

2.1givenL1L2L3algebra

The data and their first properties. By [L1] and step 1.1, $E$ is Borel, so $f = 1_{E}$ is Borel measurable. Since $0 \leq f \leq 1$ , it lies in $L^{\infty} (T, m)$ with $∥ f ∥_{\infty} \leq 1$ ; as $m (T) = 1$ , it also lies in $L^{1} (T, m)$ with $∥ f ∥_{1} \leq 1$ . Hence the Poisson integral $u = P [f]$ is defined, is complex harmonic on $D$ , and satisfies $∥ u_{r} ∥_{\infty} \leq 1$ for every $0 \leq r < 1$ .

2.2givenstep 1.1L1L2L4L5algebra

Kernel estimate near the point $q (0)$ . Let $0 \leq r < 1$ with $r \geq \frac{1}{2}$ , put $ε : = 1 - r$ , and let $η \in E$ ; write $d : = d (q (0), η)$ and choose the representative $s \in (- \frac{1}{4}, \frac{1}{4})$ of $η$ , so that $∣ s ∣ = d$ . Since $π ∣ s ∣ < \frac{π}{4} < 2$ , [L4] gives $\sin (π ∣ s ∣) \geq \frac{π}{3} ∣ s ∣$ , hence $1 - \cos (2 π s) = 2 \sin^{2} (π ∣ s ∣) \geq \frac{2 π^{2}}{9} ∣ s ∣^{2} \geq 2 d^{2}$ . By [L2], using $φ ([0]) = 1$ and $1 - 2 r \cos (2 π s) + r^{2} = (1 - r)^{2} + 2 r (1 - \cos (2 π s))$ , and using $r \geq \frac{1}{2}$ and $1 - r^{2} \leq 2 ε$ , $P (r, η) = \frac{1 - r^{2}}{1 - 2 r \cos (2 π s) + r^{2}} \leq \frac{2 ε}{ε^{2} + 2 d^{2}} .$ If moreover $η \in I_{n}$ , then $d \geq t_{n} - w_{n} \geq \frac{63}{64} t_{n}$ , so $2 d^{2} \geq t_{n}^{2}$ and $P (r, η) \leq \frac{2 ε}{ε^{2} + t_{n}^{2}}$ ; integrating this constant bound over $I_{n}$ and using $m (I_{n}) = 2 w_{n}$ from step 1.1 gives $\int_{I_{n}} P (r, η) d m (η) \leq \frac{4 ε w_{n}}{ε^{2} + t_{n}^{2}}$ .

2.3givenstep 1.1L1L2L4L5algebra

A fixed positive value near each arc. Fix $n \geq 3$ , put $r_{n} : = 1 - w_{n}$ and $z_{n} : = r_{n} φ ([t_{n}])$ , and let $J_{n} : = I_{w_{n} / 2} (q (t_{n}))$ , so that $J_{n} \subseteq I_{n}$ and $m (J_{n}) = w_{n}$ by step 1.1. Every $η \in J_{n}$ has a representative $t_{n} + σ$ with $∣ σ ∣ < \frac{1}{2} w_{n}$ , and then [L4] gives $∣ e^{2 π i σ} - 1 ∣^{2} = 2 - 2 \cos (2 π σ) = 4 \sin^{2} (π σ) \leq 4 π^{2} σ^{2}$ , so $∣ e^{2 π i σ} - 1 ∣ \leq 2 π ∣ σ ∣$ and $∣ φ (η) - z_{n} ∣ = ∣ e^{2 π i σ} - (1 - w_{n}) ∣ \leq ∣ e^{2 π i σ} - 1 ∣ + w_{n} \leq (π + 1) w_{n} .$ Since the kernel is positive and $f = 1_{E} \geq 1_{I_{n}} \geq 1_{J_{n}}$ , [L2] and [L5] give $u (z_{n}) \geq \int_{J_{n}} P (z_{n}, η) d m (η) \geq m (J_{n}) \frac{1 - r_{n}^{2}}{(π + 1)^{2} w_{n}^{2}} = \frac{2 - w_{n}}{(π + 1)^{2}} \geq \frac{1}{(π + 1)^{2}},$ where $1 - r_{n}^{2} = 2 w_{n} - w_{n}^{2}$ .

2.4givenstep 1.1L1algebra

The boundary values have no limit at $1$ . By [L1] the quotient map $q$ is continuous and $t_{n} \to 0$ , so $q (t_{n}) \to q (0)$ and $q (\frac{3}{2} t_{n}) \to q (0)$ . Step 1.1 gives $f (q (t_{n})) = 1$ while $f (q (\frac{3}{2} t_{n})) = 0$ for every $n \geq 3$ ; along the two sequences in $T$ converging to $φ ([0]) = 1$ , the values of $f$ are constantly $1$ and constantly $0$ . Hence $f$ has no limit at $φ ([0])$ .

3.1givenstep 1.1step 2.2L2L5algebra

The radial limit is zero. For $\frac{1}{2} \leq r < 1$ and $ε = 1 - r$ , [L2] and $f = 1_{E}$ give $u (r φ ([0])) = \int_{E} P (r, η) d m (η)$ ; since the arcs $I_{n}$ are pairwise disjoint with union $E$ , monotone convergence applied to the partial sums of the nonnegative functions $P (r, \cdot) 1_{I_{n}}$ gives $u (r φ ([0])) = \sum_{n \geq 3} \int_{I_{n}} P (r, η) d m (η)$ , and step 2.2 bounds this by $\sum_{n \geq 3} \frac{4 ε w_{n}}{ε^{2} + t_{n}^{2}}$ . Splitting the sum into the indices with $t_{n} \geq ε$ and those with $t_{n} < ε$ : the first part is at most $4 ε \sum_{n \geq 3} w_{n} / t_{n}^{2} = 4 ε \sum_{n \geq 3} 2^{- n} = ε$ , while the second is at most $\frac{4}{ε} \sum_{t_{n} < ε} w_{n}$ , and the indices of the second part form a tail ${n \geq N_{0}}$ with $t_{N_{0}} < ε$ and $\sum_{n \geq N_{0}} 2^{- 3 n} \leq 2 \cdot 2^{- 3 N_{0}} < 2 ε^{3}$ , so the second part is at most $8 ε^{2}$ . Therefore $u (r φ ([0])) \leq ε + 8 ε^{2}$ for $\frac{1}{2} \leq r < 1$ , and $u (r φ ([0])) \to 0$ as $r ↑ 1$ .

3.2givenstep 2.3L1L4algebra

The approach is tangential. For $n \geq 3$ , $∣ z_{n} - 1 ∣ \geq ∣ e^{2 π i t_{n}} - 1 ∣ - w_{n} = 2 \sin (π t_{n}) - w_{n}$ , and $π t_{n} \leq \frac{π}{8} < 2$ , so [L4] gives $2 \sin (π t_{n}) \geq \frac{2 π}{3} t_{n} \geq 2 t_{n}$ and hence $∣ z_{n} - 1 ∣ \geq 2 t_{n} - w_{n} \geq t_{n}$ , because $w_{n} \leq t_{n}$ . Therefore $\frac{∣ z_{n} - 1 ∣}{1 - ∣ z_{n} ∣} = \frac{∣ z_{n} - 1 ∣}{w_{n}} \geq \frac{t_{n}}{w_{n}} = 2^{2 n} \to \infty$ , while $∣ z_{n} - 1 ∣ \leq 2 π t_{n} + w_{n} \to 0$ , so $z_{n} \to 1 = φ ([0])$ . If $z_{n} \in Γ_{A} (1)$ for some $A > 1$ , then $∣ z_{n} - 1 ∣ < A (1 - ∣ z_{n} ∣) = A w_{n}$ , contradicting $∣ z_{n} - 1 ∣ \geq 2^{2 n} w_{n}$ as soon as $2^{2 n} \geq A$ ; thus for every $A > 1$ one has $z_{n} \notin Γ_{A} (1)$ for all sufficiently large $n$ .

4.1givenstep 2.1step 3.1step 2.3step 3.2step 2.4L6∎

Assembly. Step 2.1 exhibits a bounded Borel $f$ with values in ${0, 1}$ whose Poisson integral $u$ is complex harmonic; step 3.1 gives the radial limit $u (r φ ([0])) \to 0$ at $1$ , whereas step 2.3 gives $u (z_{n}) \geq (π + 1)^{- 2} > 0$ along the sequence $z_{n} \to 1$ of step 3.2, whose approach ratio $∣ z_{n} - 1 ∣ / (1 - ∣ z_{n} ∣)$ is unbounded; step 2.4 records that the boundary data themselves have no limit at $1$ . By step 3.2 each $z_{n}$ eventually lies outside every region $Γ_{A} (1)$ , so the sequence tests a path that the almost-everywhere nontangential theorem [L6] does not control; no contradiction with that theorem arises, and radial convergence at a single boundary point does not force convergence along arbitrary tangential approaches to that point.

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