3.6.2 In Harmonic Analysis
Consider , , and in (3.6.5):
Since is continuous on and is periodic with period , it admits a Fourier series representation with coefficients
so that the corresponding Fourier series is
This series may diverge. Observe that continuity of on the compact set implies uniform boundedness: such that for all (Theorem 1.2.13). Consequently, . Introducing factors with yields a convergent series:
Substituting the coefficients gives
By Theorem 2.3.2 and Theorem 2.3.6,
The summation simplifies as follows:
Substituting into (3.6.17) yields
Furthermore, by the proof of Theorem 3.6.1 (specifically (3.6.11)),
Thus, for any continuous function on , its Fourier series is Abel summable to .
We now establish that real-valued continuous functions satisfying the mean-value property are harmonic.
Theorem 3.6.2.
Let be open and continuous. Suppose for every , there exists with such that for all ,
Then is harmonic on .
Proof.
Fix arbitrarily and choose such that . Because , Theorem 3.6.1 guarantees the existence of a unique harmonic function on satisfying
with on . Define on . Then is continuous, satisfies the mean-value property, and vanishes on . By the proof of Theorem 3.4.1, which relies solely on the mean-value property, on . Thus, on , implying is harmonic at . The arbitrariness of establishes harmonicity on .