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A bounded harmonic boundary problem without uniqueness

Statement

Let $E = {a, b}$ and take the identity transition matrix $p (a, a) = p (b, b) = 1, p (a, b) = p (b, a) = 0.$ Set $A = {a}$ and prescribe the boundary value $f (a) = 1$ . For every $c \in [0, 1]$ , the bounded function $v_{c} (a) = 1$ , $v_{c} (b) = c$ satisfies $v_{c} = f$ on $A$ and $v_{c} = P v_{c}$ on $A^{c}$ , where $P q (x) : = \sum_{y \in E} p (x, y) q (y)$ for $q : E \to R$ , but under the deterministic start at $b$ , $P_{b} (T_{A} < \infty) = 0.$ Thus, without almost-sure boundary hitting, the bounded harmonic extension need not be unique. This finite witness uses no choice and remains valid when AC is assumed for the general Dirichlet theorem it illustrates.

Facts & Assumptions

Given: the two-state identity transition matrix, boundary set $A = {a}$ , and boundary value $f (a) = 1$ .

[F1]

The hitting time is $T_{A} = \inf {n \geq 0 : X_{n} \in A}$ , with the empty infimum equal to $+ \infty$ . (Hitting, return, and visit times)

[F2]

The referenced bounded Dirichlet theorem is stated under AC. (Bounded Dirichlet problem for hitting probabilities)

[F3]

The bounded Dirichlet uniqueness theorem also assumes $P_{x} (T_{A} < \infty) = 1$ for every $x \in E$ . (Bounded Dirichlet problem for hitting probabilities)

[F4]

Under AC and the all-start hitting assumption, the theorem asserts uniqueness among bounded solutions with the specified boundary values and harmonic equation on $A^{c}$ . (Bounded Dirichlet problem for hitting probabilities)

Proof

technique · construct the finite deterministic chain, calculate the harmonic equation at the sole interior state, and exhibit two different bounded solutions while the boundary is never hit from that state

1.1given

The matrix has nonnegative entries and each row sums to one. On the finite sample space $Ω = E$ with $F = 2^{E}$ , define $X_{n} (ω) = ω$ for every $n \geq 0$ . For each $x \in E$ , take the deterministic-start law $P_{x} = δ_{x}$ and the constant filtration $F_{n} = 2^{E}$ . Then $X_{n + 1} = X_{n}$ on every path, so this is a Markov chain with the displayed transition matrix. The construction uses no choice.

1.2given

For any $c \in [0, 1]$ , $v_{c}$ takes values in $[0, 1]$ , so it is bounded. Its value on $A$ is $v_{c} (a) = 1 = f (a)$ .

2.1step 1.2given

Since $A^{c} = {b}$ , the local row-sum definition of $P$ and the matrix entries give $P v_{c} (b) = p (b, a) v_{c} (a) + p (b, b) v_{c} (b) = 0 \cdot 1 + 1 \cdot c = c = v_{c} (b) .$ Thus every $v_{c}$ solves both the boundary and harmonic equations. Taking $c = 0$ and $c = 1$ gives distinct solutions, since their values at $b$ differ.

2.2F1step 1.1given

Under $P_{b} = δ_{b}$ , step 1.1 gives $X_{n} = b \notin A$ for every $n \geq 0$ . More precisely, ${n \geq 0 : X_{n} (ω) \in A}$ is empty at $ω = b$ , and $P_{b} ({b}) = 1$ ; by [F1], $T_{A} = + \infty$ almost surely and $P_{b} (T_{A} < \infty) = 0$ . In contrast, under $P_{a}$ , the initial state lies in $A$ , so $T_{A} = 0$ .

3.1F2F3step 2.2given

The referenced uniqueness theorem is stated under AC [F2] and assumes all-start almost-sure hitting [F3]. The present witness is choice-free and violates the latter assumption at $b$ , as step 2.2 shows.

4.1F4step 2.1step 2.2step 3.1given

Step 2.1 gives distinct bounded solutions, while [F4] guarantees uniqueness only under the additional hypotheses just described. Thus the counterexample does not conflict with the theorem.

5.1F1step 1.1step 1.2step 2.1step 2.2given∎

The example fixes two distinct states, so an empty or one-state space cannot instantiate it. The off-diagonal transition weights are zero, and both rows are absorbing. The endpoint $T_{A} = 0$ occurs from $a$ ; from $b$ the hitting time is infinite. The choices $c = 0$ and $c = 1$ are included and still give bounded solutions. The chain law and all calculations are explicit on a finite space, so no choice principle is used. This is a single counterexample, not an iff assertion.

Source notes

LPW, §9.2, Proposition 9.1 and its complete proof, printed pp. 117–118 (PDF pp. 132–133), proves a bounded harmonic-extension uniqueness result for an irreducible chain. Its section assumes irreducibility, which this identity matrix does not satisfy, so it is context and does not prove the counterexample. Roch, Note 24, §2, Example 24.3 and Theorem 24.4 with its first-step proof, printed/PDF pp. 3–4, discusses hitting probabilities and nonnegative exit equations; it does not state a uniqueness counterexample. The displayed two-state harmonic equations and hitting probability are calculated directly above.

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Sources

Levin, Peres and Wilmer, Markov Chains and Mixing Times, second edition (standard reference, not scraped)
Roch, Lecture Notes on Measure-Theoretic Probability Theory, Note 24 (standard reference, not scraped)