Nonabsolutely convergent Poisson integrals

Erik Talvila

Department of Mathematical and Statistical Sciences University of Alberta Edmonton AB Canada T6G 2E2 etalvila@math.ualberta.ca

Abstract. If a function f has finite Henstock integral on the boundary of the unit disk of ℝ^{2} then its Poisson integral exists for |z| < 1 and is o((1 − |z|)^{−1}) as |z| → 1^{−}. It is shown that this is the best possible uniform pointwise estimate. For an L^{1} measure the best estimate is O((1 − |z|)^{−1}). AMS (MOS) subject classification: 26A39, 31A20

In this paper we consider estimates of Poisson integrals on the unit circle with respect to Alexiewicz and L^{p} norms. Define the open disk in ℝ^{2} as D := {re^{iθ} | 0 ≤ r < 1, −π < θ ≤ π} and let the unit circle T be its boundary. Let f: T → ℝ. The Poisson integral of f with respect to Lebesgue measure is

P [f](re^{iθ}) = \frac{(1 − r^{2})}{2π} ∫_{ϕ=−π}^{π} \frac{f(ϕ) dϕ}{1 − 2r cos(ϕ − θ) + r^{2}} .

For the Poisson integral of f with respect to measure μ we write P [f, μ]. Since T has no end points, an appropriate form of the Alexiewicz norm of f is ∥f∥ := sup_{I⊆T} |∫_{I} f| where I is an interval in T. Hence, we can have I = [α, β] where α, β ∈ ℝ and 0 ≤ β − α ≤ 2π. The Alexiewicz norm was introduced in [1]. The variation of f on T is sup ∑_{i=1}^{N} |f(x_{i−1})−f(x_{i})| where the supremum is taken over all finite sets of disjoint intervals {(x_{i−1}, x_{i})}_{i=1}^{N} in [−π, π]. We denote the variation of f over I ⊆ T as V_{I}f.

The following results are well known (see [4]). Suppose that 1 ≤ p ≤ ∞ and f ∈ L^{p}(T). If e^{iθ_{0}} ∈ T and z ∈ D, we say that z → e^{iθ_{0}} nontangentially if there is 0 ≤ α < π/2 such that z → e^{iθ_{0}} with z remaining in the cone K_{α}(e^{iθ_{0}}) := {ζ ∈ D : | arg(ζ − e^{iθ_{0}}) − θ_{0}| < α}. Write u_{r}(θ) = P [f](re^{iθ}). Then

1. u_{r} is harmonic in D

2. ∥u_{r}∥_{p} ≤ ∥f∥_{p} for all 0 ≤ r < 1

3. If 1 ≤ p < ∞ then ∥u_{r} − f∥_{p} → 0 as r → 1^{−}

4. u_{r} → f almost everywhere on T as r → 1 nontangentially in D.

We examine analogues of these results when f is Henstock integrable. All the results also hold when we use the wide Denjoy integral.

Necessary and sufficient for the existence of P [f] as a Henstock integral on D is that f be integrable, i.e., the Henstock integral ∫_{−π}^{π}f is finite. This is because the kernel (1−r^{2})/[1−2r cos(ϕ−θ)+r^{2}] is bounded away from 0 and is of bounded variation in ϕ for each re^{iθ} ∈ D. In [2], integration by parts was used to show that we can differentiate under the integral sign. This in turn shows that P [f] is harmonic in D and that P [f] → f nontangentially, almost everywhere in T (4. above). In [3], Theorem 4, p. 238, necessary and sufficient conditions were given for determining when a function that is harmonic on D is the Poisson integral of a Henstock integrable function. Corresponding results when ∥u_{r}∥_{p} are uniformly bounded have been known for some time ([4], Theorem 11.30).

Our first result is to show that P [f](re^{iθ}) = o(1/(1 − r)) as r → 1^{−}. That is, sup_{θ∈[−π,π]}(1 − r)|P [f](re^{iθ})| → 0 as r → 1^{−}. Thus, the manner of approach to the boundary is unrestricted. This same estimate was obtained for Lebesgue integrable functions in [6]. We show it is the best possible pointwise estimate under our minimal existence hypothesis.

Theorem 1 i) Let f:T → ℝ. If f is integrable then P [f](re^{iθ}) = o(1/(1 − r)) as r → 1^{−}. This estimate is sharp in the sense that if ψ : D → ℝ and ψ(re^{iθ}) = o(1/(1 − r)) then there is an integrable function f such that P [f] ≠ o(ψ). ii) Let μ be a positive measure on T. If f is in L^{1}(μ) then P [f, dμ](re^{iθ}) = O(1/(1 − r)). This estimate is sharp in the same sense as in i).

Proof: Let Φ_{r}(ϕ) := (1 − r)^{2}/(1 − 2r cos ϕ + r^{2}) with Φ_{1}(0) := 1 and f_{θ}(ϕ) := f((ϕ + θ) mod 2π). Then

(1 − r)P [f](re^{iθ}) = \frac{(1 + r)}{2π} ∫_{ϕ=−π}^{π} f_{θ}(ϕ) Φ_{r}(ϕ) dϕ.

Write

\frac{2π(1 − r)P [f](re^{iθ})}{1 + r} = ∫_{|ϕ|<δ} f_{θ}(ϕ)Φ_{r}(ϕ) dϕ + ∫_{δ<|ϕ|<π}f_{θ}(ϕ)Φ_{r}(ϕ) dϕ.

Let F_{θ}(ϕ) = ∫_{−δ}^{ϕ} f_{θ} and integrate by parts. Then

| ∫_{|ϕ|<δ} f_{θ}(ϕ)Φ_{r}(ϕ) dϕ| = |F_{θ}(δ)Φ_{r}(δ) − ∫_{ϕ=−δ}^{δ} F_{θ} dΦ_{r}(ϕ)|

≤ |∫_{θ−δ}^{θ+δ} f| (1 + V_{T} Φ_{r}) .  (1)

But V_{T} Φ_{r} = 8r/(1 + r)^{2} ≤ 2. And, since the integral is continuous with respect to its limits of integration, by taking δ > 0 small enough we can make the right side of (1) as small as we please.

Letting G_{θ}(ϕ) := ∫_{δ}^{ϕ} f_{θ}, we have

| ∫_{ϕ=δ}^{π} f_{θ}(ϕ)Φ_{r}(ϕ) dϕ| = |G_{θ}(π)Φ_{r}(π) − ∫_{ϕ=δ}^{π} G_{θ} dΦ_{r}(ϕ)|

≤ ∥f∥ |(\frac{1 − r}{1 + r})^{2} + \frac{(1 − r)^{2}}{1 − 2r cos δ + r^{2}} |
→ 0 as r → 1.

Similarly, ∫_{−π}^{−δ} f_{θ}(ϕ)Φ_{r}(ϕ) dϕ → 0 as r → 1.

To prove sharpness, suppose ψ: D → ℝ is given. It suffices to show that P [f](r_{n}e^{iθ_{n}}) ≠ o(ψ(r_{n}e^{iθ_{n}})) for some sequence {r_{n}e^{iθ_{n}}} ∈ D with r_{n} → 1^{−}. Take r_{n}e^{iθ_{n}} → 1 and θ_{n} ↓ 0. Let a_{n} = |ψ(r_{n}e^{iθ_{n}})| and let {α_{n}} and {f_{n}} be sequences of positive numbers. Define

f(ϕ) = \{ \begin{matrix}f_{n}, & |ϕ − θ_{n}| ≤ α_{n} for some n \\ 0, & otherwise. \end{matrix}

For n = 1, 2, 3, ⋯ take 0 < α_{n} ≤ π−θ_{n} and small enough so that the intervals (θ_{n}−α_{n}, θ_{n}+α_{n}) are disjoint. This will be so if α_{n} ≤ \frac{1}{2} min(θ_{n−1}−θ_{n}, θ_{n}−θ_{n+1}) (θ_{0} := π). Now,

πP [f](r_{n}e^{iθ_{n}}) = (1 + r_{n})(1 − r_{n}) ∑_{k=1}^{∞} f_{k} ∫_{ϕ=θ_{k}−α_{k}}^{θ_{k}+α_{k}}\frac{dϕ}{r_{n}^{2} − 2r_{n} cos(θ_{n} − ϕ) + 1}

≥ \frac{2(1 + r_{n})(1 − r_{n})f_{n} α_{n}}{r_{n}^{2} − 2r_{n} cos(α_{n}) + 1}
≥ \frac{2(1 + r_{n})(1 − r_{n})f_{n} α_{n}}{(1 − r_{n})^{2} + r_{n}α_{n}^{2}} .

Hence, taking α_{n} = min(\frac{1}{2}(θ_{n−1} − θ_{n}), \frac{1}{2}(θ_{n} − θ_{n+1}), 1 − r_{n}) and f_{n} = π(1 − r_{n})a_{n}/α_{n} gives P [f](r_{n}e^{iθ_{n}}) ≥ a_{n}. And, f ∈ L^{1} if ∑ f_{k} α_{k} = π ∑(1−r_{k})a_{k} < ∞. Since (1 − r_{k})a_{k} → 0 there is a subsequence {(1 − r_{n})a_{n}}_{n∈I} defined by an unbounded index set I ⊂ ℕ such that ∑_{k∈I} a_{k} < ∞. Then, f ∈ L^{1} and P [f](r_{n}e^{iθ_{n}}) ≥ |ψ(r_{n}e^{iθ_{n}})| for all n ∈ I.

For ii), let f ∈ L^{1}(dμ). Then

|P [f, μ](r, θ)| ≤ \frac{1 − r^{2}}{2π(1 − r)^{2}} ∫_{−π}^{π} | f | dμ = O (\frac{1}{1 − r}) .

The estimate is realised with the Dirac measure, i.e., if ϕ_{0} ∈ [−π, π] then (1 − r)P [1, δ_{ϕ_{0}}](r, ϕ_{0}) = (1 + r)/(2π) → 1/π as r → 1^{−}. ■

In part i), the sharpness is in fact realised with data that is positive (and hence L^{1}). The electrostatic interpretation of ii) is a unit charge at z = 1.

The analogues of properties 2. and 3. are now considered for the Alexiewicz norm.

Theorem 2 Let f : T → ℝ be integrable. For re^{iθ} ∈ D define u_{r}(θ) := P [f](re^{iθ}). Then

i) ∥u_{r}∥ ≤ ∥f∥ for all 0 ≤ r < 1

ii) ∥u_{r} − f∥ → 0 as r → 1^{−}

iii) In ii), the decay of ∥u_{r} − f∥ can be arbitrarily slow.

Proof: i) Let α ∈ ℝ and 0 < β − α ≤ 2π. Then

∫_{θ=α}^{β} u_{r}(θ) dθ = ∫_{θ=α}^{β} \frac{(1 − r^{2})}{2π} ∫_{ϕ=−π}^{π} \frac{f(ϕ) dϕ dθ}{1 − 2r cos(ϕ − θ) + r^{2}} .  (2)

If r = 0 it is clear that ∥u_{0}∥ ≤ ∥f∥ so assume 0 < r < 1. By Theorem 57 (p. 58) or Theorem 58 (p. 60) in [3] or by [5] we can interchange the orders of integration in (2). Let v_{r}(θ) = P [χ_{[α,β]}](re^{iθ}). Then

∫_{θ=α}^{β} u_{r}(θ) dθ = ∫_{ϕ=−π}^{π} f(ϕ)v_{r}(ϕ) dϕ.

If β − α = 2π then v_{r} = 1 on T and the result is immediate. Now assume 0 < β − α < 2π. For fixed r the function v_{r} has one maximum, at ϕ_{1} := (α + β)/2, and one minimum, at ϕ_{2} := ϕ_{1} + π. Now use the Bonnet form of the Second Mean Value Theorem for integrals ([3], p. 34) to write

∫_{θ=α}^{β} u_{r}(θ) dθ = ∫_{ϕ=ϕ_{1}}^{ϕ_{2}} f(ϕ)v_{r}(ϕ) dϕ + ∫_{ϕ=ϕ_{2}}^{ϕ_{2}+π}f(ϕ)v_{r}(ϕ) dϕ

= v_{r}(ϕ_{1}) ∫_{ϕ_{1}}^{ξ_{1}}f + v_{r}(ϕ_{1}) ∫_{ξ_{2}}^{ϕ_{2}+π} f
= v_{r}(ϕ_{1}) ∫_{ξ_{2}}^{ξ_{1}}f

where ϕ_{1} < ξ_{1} < ϕ_{2} and ϕ_{2} < ξ_{2} < ϕ_{2} + π. Now,

|∫_{α}^{β} u_{r}| ≤ max_{θ∈[−π,π]} v_{r}(θ) |∫_{ξ_{2}}^{ξ_{1}} f|
≤ ∥f∥.

It now follows that ∥u_{r}∥ ≤ ∥f∥.

ii) Let α ∈ ℝ and 0 < β − α ≤ 2π. Suppose ϵ > 0 is given. There are functions a: (0, 1) → [−π, π] and b: (0, 1) → [−π, π] such that for each 0 < r < 1 we have

∥u_{r} − f∥ ≤ | ∫_{θ=a(r)}^{b(r)}[u_{r}(θ) − f(θ)] dθ| + ϵ.

Let r_{n} ↑ 1. There is then a subsequence {r_{n_{m}}} on which a converges, say a(r_{n_{m}}) → α as r_{n_{m}} → 1. By taking a piecewise linear function that agrees with a at the points r_{n_{m}}, we can assume a is continuous and has limit α. Similarly, we can assume b is continuous and has limit β where 0 < β − α < 2π. Note that

∫_{θ=a(r)}^{b(r)} u_{r}(θ) dθ − ∫_{θ=α}^{β} u_{r}(θ) dθ = ∫_{θ=−π}^{π} f(θ)P [χ_{[a(r),b(r)]} − χ_{[α,β]}](re^{iθ}) dθ. (3)

The variation of P [χ_{[a(r),b(r)]} − χ_{[α,β]}](re^{iθ}) over θ ∈ T is at most 4. And, P [χ_{[a(r),b(r)]} − χ_{[α,β]}](re^{iθ}) → 0 as r → 1 for θ ≠ α, β. We can bring this limit under the integral sign and it follows that both sides of (3) tend to 0 as r → 1. Since integrals are continuous with respect to their limits of integration we will have

∥u_{r} − f∥ ≤ | ∫_{θ=α}^{β} [u_{r}(θ) − f(θ)] dθ| + 2ϵ,

for r close enough to 1. And,

∫_{θ=α}^{β} [u_{r}(θ) − f(θ)] dθ = ∫_{ϕ=−π}^{π} f(ϕ)v_{r}(ϕ) dϕ − ∫_{ϕ=α}^{β} f(ϕ) dϕ

= ∫_{ϕ=−π}^{π} f(ϕ)ψ_{r}(ϕ) dϕ  (4)

where ψ_{r} := v_{r} − χ_{[α,β]}.

Now, ψ_{r} has variation at most 2. Hence, it is of bounded variation, uniformly with respect to 0 ≤ r ≤ 1, α ∈ ℝ and β ∈ ℝ such that 0 ≤ β − α ≤ 2π. And,

ψ_{r}(ϕ) → \{ \begin{matrix}0, & ϕ ≠ α, β \\ −1/2, & ϕ = α, β. \end{matrix}

Taking the limit r → 1^{−} inside the integral (4) now gives ∥u_{r} − f∥ → 0 as r → 1^{−}.

iii) Let f be positive on (0, 1) and vanish elsewhere. Then u_{r} is positive for 0 ≤ r < 1. We then have

∥u_{r} − f∥ ≥ ∫_{ϕ=−π}^{0} u_{r}(ϕ) dϕ

= ∫_{ϕ=−π}^{0} \frac{1 − r^{2}}{2π} ∫_{θ=0}^{1} \frac{f(θ) dθ dϕ}{1 − 2r cos(θ − ϕ) + r^{2}}

= ∫_{θ=0}^{1} f(θ)P [χ_{[−π,0]}](re^{iθ}) dθ.

Now, as r → 1

P [χ_{[−π,0]}](re^{iθ}) → \{ \begin{matrix}0, & 0 < θ < π \\ 1/2, & θ = −π, 0 \\ 1, & −π < θ < 0. \end{matrix}

But, the convergence is not uniform. Let a decay rate be given by A:(0, 1) → (0, 1/2), where A(r) decreases to 0 as r increases to 1. By keeping θ close enough to 0 we can keep P [χ_{[−π,0]}](re^{iθ}) bounded away from 0 for all r. To see this, write ρ := (1 + r)/(1 − r). Then

∥u_{r} − f∥ ≥ ∫_{θ=0}^{1−r}f(θ)P [χ_{[−π,0]}](re^{iθ}) dθ

= \frac{1}{π} ∫_{θ=0}^{1−r}f(θ) \{\frac{π}{2} − arctan [ρ tan (\frac{θ}{2})] + arctan [\frac{1}{ρ} tan (\frac{θ}{2})]\} dθ

≥ ∫_{θ=0}^{1−r}f(θ) \{\frac{1}{2} − \frac{1}{π} arctan [ρ tan (\frac{θ}{2})]\} dθ

≥ ∫_{θ=0}^{1−r}f(θ) \{\frac{1}{2} − \frac{ρθ}{2π} \} dθ

≥ (\frac{1}{2} − \frac{1}{π} ) ∫_{θ=0}^{1−r}f(θ)dθ.

We can now let

f(θ) := \{ \begin{matrix}− (\frac{1}{2} − \frac{1}{π})^{−1} A′(1 − θ), & 0 < θ < 1 \\ 0, & otherwise. \end{matrix}

And,

∥u_{r} − f∥ ≥ − ∫_{θ=0}^{1−r}A′(1 − θ) dθ = A(r). ■

Remarks.

1. We have equality in i) when f is of one sign.

2. The triangle inequality and ii) show that ∥u_{r}∥ → ∥f∥ as r → 1.

3. In iii), the decay of ∥u_{r} − f∥ can be arbitrarily rapid. Take f to be constant!

4. The same proof shows that we can choose f ∈ L^{1} to make ∥u_{r} − f∥_{1} tend to 0 arbitrarily slowly. Jensen’s inequality then shows the same holds true for ∥u_{r} − f∥_{p} for some f ∈ L^{p}, for each 1 ≤ p < ∞.

We now look at the interplay between the Alexiewicz and L^{p} norms. In Theorem 1 we saw that P [f](re^{iθ}) has the same best pointwise estimate o(1/(1 − r) when f is Henstock integrable or in L^{1}. The L^{∞} norm is thus too coarse for it to show a size difference. However, for 1 ≤ p < ∞ the L^{p} norms of P [f] are substantially larger when P [f] can converge conditionally.

Theorem 3 Let f : T → ℝ be integrable. For re^{iθ} ∈ D define u_{r}(θ) := P [f](re^{iθ}). Then ∥u_{r}∥_{p} = o(1/(1 − r)) for 1 ≤ p < ∞.

Proof: From Theorem 1 we can write u_{r}(θ) = w_{r}(θ)/(1 − r) where sup_{θ∈[−π,π]} |w_{r}(θ)| → 0 as r → 1. And, w_{r} is periodic and real analytic on [−π, π] for each 0 ≤ r < 1. Let 1 ≤ p < ∞. Then

∥u_{r}∥_{p} = \frac{1}{1 − r} [ ∫_{θ=−π}^{π} |w_{r}(θ)|^{p} dθ]^{1/p}

≤ \frac{(2π)^{1/p}}{1 − r} sup_{θ∈[−π,π]} |w_{r}(θ)|.

Hence, ∥u_{r}∥_{p} = o(1/(1 − r)) as r → 1. ■

It is not known at this time whether or not this estimate is sharp. However, an example shows that for each 0 < α < 1 and 1 ≤ p < ∞ there is an integrable function f so that lim sup ||u_{r}||_{p}(1 − r)^{α} = ∞ as r → 1.

References

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[2] M. Benedicks and W.F. Pfeffer, The Dirichlet problem with Denjoy– Perron integrable boundary condition, Canad. Math. Bull. 28 (1985) 113–119.

[3] V.G. Čelidze and A.G. Džvaršeǐšvili, The theory of the Denjoy integral and some applications (trans. P.S. Bullen), Singapore, World Scientific, 1989.

[4] W. Rudin, Real and complex analysis, New York, McGraw-Hill, 1987.

[5] E. Talvila, Limits and Henstock integrals of products, Real. Anal. Exchange 25 (1999/00) 907-918.

[6] F. Wolf, The Poisson integral. A study in the uniqueness of harmonic functions, Acta. Math. 74 (1941) 65-100.