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Fall 2009 - Problem 6

Banach-Alaoglu, Poisson kernel, Riesz representation, weak convergence

The Poisson kernel for 0≤ρ<10 \leq \rho < 1 is the 2π2\pi periodic function on R\R defined by

Pρ(θ)=Re⁡(1+ρeiθ1−ρeiθ).P_\rho\p{\theta} = \Re\p{\frac{1 + \rho e^{i\theta}}{1 - \rho e^{i\theta}}}.

For functions hh continuous on and harmonic inside the closed disc of radius RR about the origin, one has (you need not prove this)

h(reiη)=12π∫02πPr/R(η−θ)h(Reiθ) dθ.h\p{re^{i\eta}} = \frac{1}{2\pi} \int_0^{2\pi} P_{r/R}\p{\eta - \theta} h\p{Re^{i\theta}} \,\diff\theta.

Assume hh is harmonic and positive on the open unit disc D={z∈C∣∣z∣<1}D = \set{z \in \C \mid \abs{z} < 1}. Prove that there exists a poistive Borel measure μ\mu on [0,2π]\br{0, 2\pi} such that for all reiθ∈Dre^{i\theta} \in D one has

h(reiη)=∫02πPr(η−θ) dμ(θ).h\p{re^{i\eta}} = \int_0^{2\pi} P_r\p{\eta - \theta} \,\diff\mu\p{\theta}.
Solution.

For each 0<R<10 < R < 1, we have the measure

dμR=h(Reiθ)2π dθ.\diff\mu_R = \frac{h\p{Re^{i\theta}}}{2\pi} \,\diff\theta.

If 0≤r<10 \leq r < 1, then Rr<R<1Rr < R < 1, so h(Rz)h\p{Rz} is harmonic on DD and continuous on ∂D\partial{D}. Thus, by the Poisson kernel representation formula, we have for any η∈[0,2π]\eta \in \br{0, 2\pi} that

h(Rreiη)=∫02πPr(η−θ) dμR(θ).h\p{Rre^{i\eta}} = \int_0^{2\pi} P_r\p{\eta - \theta} \,\diff\mu_R\p{\theta}.

By Riesz representation, we may view {μR}R⊆C([0,2π])∗\set{\mu_R}_R \subseteq C\p{\br{0, 2\pi}}^\ast. Moreover,

∥μR∥=12π∫02πh(Reiθ) dθ=h(0),\norm{\mu_R} = \frac{1}{2\pi} \int_0^{2\pi} h\p{Re^{i\theta}} \,\diff\theta = h\p{0},

by the mean value property for harmonic functions. Thus, by Banach-Alaoglu, {μR}R\set{\mu_R}_R admits a weak-* convergent sequence {μRn}n\set{\mu_{R_n}}_n, where Rn↗1R_n \nearrow 1, i.e., there exists a finite, positive Borel measure μ∈C([0,2π])∗\mu \in C\p{\br{0, 2\pi}}^\ast such that

∫02πf(θ) dμRn(θ)→n→∞∫02πf(θ) dμ(θ)\int_0^{2\pi} f\p{\theta} \,\diff\mu_{R_n}\p{\theta} \xrightarrow{n\to\infty} \int_0^{2\pi} f\p{\theta} \,\diff\mu\p{\theta}

for any f∈C([0,2π])f \in C\p{\br{0, 2\pi}}. In particular, setting f(θ)=Pr(η−θ)f\p{\theta} = P_r\p{\eta - \theta}, we have for all n≥1n \geq 1

h(Rnreiη)=∫02πPr(η−θ) dμRn.h\p{R_nre^{i\eta}} = \int_0^{2\pi} P_r\p{\eta - \theta} \,\diff\mu_{R_n}.

Taking n→∞n \to \infty, we get Rn→1R_n \to 1, so Rnreiη→reiη∈DR_nre^{i\eta} \to re^{i\eta} \in D where hh is harmonic, hence continuous, so by weak-* convergence,

h(reiη)=∫02πPr(η−θ) dμ(θ).h\p{re^{i\eta}} = \int_0^{2\pi} P_r\p{\eta - \theta} \,\diff\mu\p{\theta}.