Measure, Integration & Real Analysis, 2021a

Section 11A Fourier Series and Poisson Integral 345
11.11 Definition P r f
For f ∈ L 1 (∂D) and 0 ≤ r < 1, define P r f : ∂D → C by
(P r f )(z) =
∞
∑ r |n| ̂f (n)z n .
n=−∞
No convergence problems arise in the series above because
∣ r
|n| ̂f (n)z
n ∣ ∣ ≤‖f ‖ 1 r |n|
for each z ∈ ∂D, which implies that
∞
∑ |r |n| ̂f (n)z n 1 + r
|≤‖f ‖ 1
n=−∞
1 − r < ∞.
Thus for each r ∈ [0, 1), the partial sums of the series above converge uniformly on
∂D, which implies that P r f is a continuous function from ∂D to C (for r = 0 and
n = 0, interpret the expression 0 0 to be 1).
Let’s unravel the formula in 11.11. Iff ∈ L 1 (∂D), 0 ≤ r < 1, and z ∈ ∂D, then
11.12
(P r f )(z) =
=
∫
=
∞
∑ r |n| ̂f (n)z
n
n=−∞
∞ ∫
∑ r |n|
n=−∞ ∂D
∂D
f (w)w n dσ(w)z n
f (w)( ∞
∑ r |n| (zw) n) dσ(w),
n=−∞
where interchanging the sum and integral above is justified by the uniform convergence
of the series on ∂D. To evaluate the sum in parentheses in the last line above,
let ζ ∈ ∂D (think of ζ = zw in the formula above). Thus (ζ) −n =(ζ) n and
∞
∑ r |n| ζ n =
n=−∞
∞
∑ (rζ) n +
n=0
∞
∑ (rζ) n
n=1
= 1
1 − rζ + rζ
1 − rζ
=
(1 − rζ)+(1 − rζ)rζ
|1 − rζ| 2
11.13
= 1 − r2
|1 − rζ| 2 .
Motivated by the formula above, we now make the following definition. Notice
that 11.11 uses calligraphic P, while the next definition uses italic P.
Measure, Integration & Real Analysis, by Sheldon Axler