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The disc Poisson integral can miss the assigned value at a jump

Statement refuted

Let $g : \partial D \to {0, 1}$ be $g (e^{i t}) = 1$ for $0 < t < π$ and $g (e^{i t}) = 0$ for $π \leq t \leq 2 π$ , so that $g (1) = 0$ . For $0 \leq r < 1$ define its bounded-data Poisson integral by $U_{g} (r) = (2 π)^{- 1} \int_{0}^{2 π} P_{r} (t) g (e^{i t}) d t$ . Then $U_{g} (r) = 1 / 2$ for every $r$ , so $U_{g} (r) \to 1 / 2 \neq g (1)$ as $r ↑ 1$ .

Facts & Assumptions

Given: the unit circle boundary datum $g$ above and the disc Poisson kernel.

[F1]

For $z = r e^{i ϕ} \in D$ and $t \in R$ the Poisson kernel of the unit disc is $P (z, e^{i t}) = (1 - ∣ z ∣^{2}) / ∣ e^{i t} - z ∣^{2}$ ; writing $z = r e^{i ϕ}$ with $0 \leq r < 1$ gives $P (z, e^{i t}) = P_{r} (t - ϕ)$ with $P_{r} (θ) = (1 - r^{2}) / (1 - 2 r \cos θ + r^{2})$ (The Poisson kernel on the unit disc).

[F2]

For $0 \leq r < 1$ the kernel $P_{r} (θ) = (1 - r^{2}) / (1 - 2 r \cos θ + r^{2})$ satisfies $P_{r} (θ) > 0$ for every $θ$ and $(2 π)^{- 1} \int_{0}^{2 π} P_{r} (θ) d θ = 1$ (The Poisson kernel is positive, has total mass one, and concentrates at a boundary point).

Counterexample

technique · direct

1.1givenF1F2

Define $g (e^{i t}) : = 1$ for $0 < t < π$ , $g (e^{i t}) : = 0$ for $π \leq t \leq 2 π$ , and $U_{g} (r) : = (2 π)^{- 1} \int_{0}^{2 π} P_{r} (t) g (e^{i t}) d t$ for $0 \leq r < 1$ , with $P_{r}$ as in [F1]; this integral is finite because $g$ is bounded and the kernel is continuous on the compact circle. Since $g (e^{i t})$ vanishes on $[π, 2 π]$ and equals one on $(0, π)$ , $U_{g} (r) = (2 π)^{- 1} \int_{0}^{π} P_{r} (t) d t$ . Also $g (1) = g (e^{i 0}) = 0$ , because $0$ is not an interior point of $(0, π)$ .

2.1givenstep 1.1F1algebra

Reflection symmetry. For every $t$ we have $\cos (2 π - t) = \cos t$ , so [F1] gives $P_{r} (2 π - t) = P_{r} (t)$ ; the substitution $t \mapsto 2 π - t$ maps $(0, π)$ onto $(π, 2 π)$ and preserves the Lebesgue measure. Hence $\int_{0}^{π} P_{r} (t) d t = \int_{π}^{2 π} P_{r} (t) d t$ .

3.1step 1.1step 2.1F2algebra

Normalization. By [F2], $(2 π)^{- 1} \int_{0}^{2 π} P_{r} (θ) d θ = 1$ ; splitting the integral at $π$ and using step 2.1, $1 = 2 \cdot (2 π)^{- 1} \int_{0}^{π} P_{r} (t) d t$ . Therefore $U_{g} (r) = (2 π)^{- 1} \int_{0}^{π} P_{r} (t) d t = 1 / 2$ for every $0 \leq r < 1$ .

4.1step 1.1step 3.1F1∎

Failure at the jump. The value $U_{g} (r) = 1 / 2$ is independent of $r$ , so $U_{g} (r) \to 1 / 2$ as $r ↑ 1$ , while the datum assigns $g (1) = 0$ at the boundary point $e^{i 0} = 1$ ; thus the Poisson integral of a bounded boundary function need not recover the assigned value at a discontinuity, and only continuity of the datum at the point would force it.

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Sources

Per Kristen Jakobsen, An Introduction to Partial Differential Equations (2019) (standard reference, not scraped)
Leon Simon, Lectures on PDE (2015 rough draft) (standard reference, not scraped)
Thomas Schmidt, Partial Differential Equations I (2026) (standard reference, not scraped)