Notes on Complex Analysis
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§3.6.2 The Poisson Integral Formula in Harmonic Analysis

3.6.2 The Poisson Integral Formula in Harmonic Analysis

Consider 𝑅=1, 𝜁=ei𝜏, and 𝑧=𝜌ei𝜃 in (3.6.5)(3.6.5)(3.6.5):

𝑢(𝑧)=12π∫02π𝑢(𝜁)1−|𝑧|2|𝜁−𝑧|2d𝜏=12π∫02π(1−𝜌2)𝑢(ei𝜏)d𝜏(ei𝜏−𝜌ei𝜃)(e−i𝜏−𝜌e−i𝜃)=12π∫02π(1−𝜌2)𝑢(ei𝜏)d𝜏1+𝜌2−2𝜌cos(𝜃−𝜏).(3.6.2.1)

Since 𝑢(𝑧) is continuous on 𝜕𝔻 and 𝑢(ei𝜃) is periodic with period 2π, it admits a Fourier series representation with coefficients

𝑎𝑛=12π∫02π𝑢(ei𝜏)e−i𝑛𝜏d𝜏,(3.6.2.2)

so that the corresponding Fourier series is

∑𝑛=−∞∞𝑎𝑛ei𝑛𝜃.

This series may diverge. Observe that continuity of 𝑢 on the compact set 𝜕𝔻 implies boundedness: ∃𝑀>0 such that |𝑢(ei𝜃)|≤𝑀 for all 𝜃 (Theorem 1.2.22Theorem 1.2.22Theorem 1.2.22). Consequently, |𝑎𝑛|≤𝑀. Introducing factors 𝜌|𝑛| with |𝜌|<1 yields a convergent series:

∑𝑛=−∞∞𝑎𝑛ei𝑛𝜃𝜌|𝑛|,|∑𝑛=−∞∞𝑎𝑛ei𝑛𝜃𝜌|𝑛||≤∑𝑛=−∞∞|𝑎𝑛|𝜌|𝑛|≤𝑀1+|𝜌|1−|𝜌|.

Substituting the coefficients gives

∑𝑛=−∞∞𝑎𝑛ei𝑛𝜃𝜌|𝑛|=∑𝑛=−∞∞(12π∫02π𝑢(ei𝜏)e−i𝑛𝜏d𝜏)ei𝑛𝜃𝜌|𝑛|=12π∑𝑛=−∞∞∫02π𝜌|𝑛|𝑢(ei𝜏)ei𝑛(𝜃−𝜏)d𝜏.

By Theorem 2.3.3Theorem 2.3.3Theorem 2.3.3 and Theorem 2.3.7Theorem 2.3.7Theorem 2.3.7,

12π∑𝑛=−∞∞∫02π𝜌|𝑛|𝑢(ei𝜏)ei𝑛(𝜃−𝜏)d𝜏=12π∫02π𝑢(ei𝜏)∑𝑛=−∞∞𝜌|𝑛|ei𝑛(𝜃−𝜏)d𝜏.(3.6.2.3)

The summation simplifies as follows:

∑𝑛=−∞∞𝜌|𝑛|ei𝑛(𝜃−𝜏)=∑𝑛=0∞𝜌𝑛ei𝑛(𝜃−𝜏)+∑𝑛=1∞𝜌𝑛e−i𝑛(𝜃−𝜏)=1+2∑𝑛=1∞𝜌𝑛cos(𝑛(𝜃−𝜏))=1+2ℜ𝔢(∑𝑛=1∞𝜌𝑛ei𝑛(𝜃−𝜏))=1+2ℜ𝔢(𝜌ei(𝜃−𝜏)1−𝜌ei(𝜃−𝜏))=1−𝜌21+𝜌2−2𝜌cos(𝜃−𝜏).

Substituting into (3.6.2.3)(3.6.2.18)(3.6.2.3) yields

∑𝑛=−∞∞𝑎𝑛ei𝑛𝜃𝜌|𝑛|=12π∫02π(1−𝜌2)𝑢(ei𝜏)1+𝜌2−2𝜌cos(𝜃−𝜏)d𝜏=𝑢(𝜌ei𝜃).

Furthermore, by the proof of Theorem 3.6.1.1Theorem 3.6.1.1Theorem 3.6.1.1 (specifically (3.6.1.2)(3.6.1.12)(3.6.1.2)),

lim𝜌→1−∑𝑛=−∞∞𝑎𝑛ei𝑛𝜃𝜌|𝑛|=𝑢(ei𝜃).

Thus, for any continuous function 𝑢 on 𝜕𝔻, its Fourier series is Abel summable to 𝑢.

We now establish that real-valued continuous functions satisfying the local mean-value property are harmonic. Here, local means that for every 𝑧0∈𝑈, there is a disk 𝐷(𝑧0,𝜆)‾⋐𝑈 such that the identity holds on every concentric disk 𝐷(𝑧0,𝜀)‾ with 0<𝜀≤𝜆. As noted above, this use of local is informal terminology adopted for clarity rather than necessarily standard classical terminology.

Theorem 3.6.2.1.

Let 𝑈⊆ℂ be an open region and 𝑓:𝑈→ℝ continuous. Suppose for every 𝑧0∈𝑈, there exists 𝜆>0 with 𝐷(𝑧0,𝜆)‾⋐𝑈 such that for all 0<𝜀≤𝜆,

𝑓(𝑧0)=12π∫02π𝑓(𝑧0+𝜀ei𝑡)d𝑡.
Then 𝑓 is harmonic on 𝑈.
Proof.

Fix 𝑧0∈𝑈 arbitrarily and choose 𝜆>0 such that 𝐷(𝑧0,𝜆)‾⋐𝑈. Because 𝑓∈𝐶(𝜕𝐷(𝑧0,𝜆)), Theorem 3.6.1.1Theorem 3.6.1.1Theorem 3.6.1.1 guarantees the existence of a unique harmonic function 𝑢 on 𝐷(𝑧0,𝜆) satisfying 𝑢≡𝑓 on 𝜕𝐷(𝑧0,𝜆) with

𝑢(𝑧)=∫02π𝑓(𝜁)𝑃(𝜁,𝑧)d𝜏.
Define 𝜓=𝑓−𝑢 on 𝐷(𝑧0,𝜆)‾. Then 𝜓 is continuous, satisfies the local mean-value property in 𝐷(𝑧0,𝜆), and vanishes on 𝜕𝐷(𝑧0,𝜆). By Theorem 3.4.3Theorem 3.4.3Theorem 3.4.3, 𝜓≡0 on 𝐷(𝑧0,𝜆)‾. Thus, 𝑓≡𝑢 on 𝐷(𝑧0,𝜆)‾, implying 𝑓 is harmonic at 𝑧0. The arbitrariness of 𝑧0 establishes harmonicity on 𝑈. ⁠□