Exercise 1.6.12 (Rising sun inequality I)

Let f : R → R be an absolutely integrable function, and let f∗ : R → R be the one-sided signed Hardy-Littlewood maximal function

f∗(x) := sup ⁡ h>0 1 h∫ [x,x+h]fdm.

Establish the rising sun inequality

λm({f∗(x) > λ}) ≤∫ {x:f∗(x)>λ}fdm

for all real λ (note here that we permit λ to be zero or negative), and show that this inequality implies Lemma 1.6.16.

Answers

We can rewrite the theorem assertion as

∫ f∗−λ>0(f − λ)dμ ≥ 0.

Since f∗− λ = (f − λ)∗, we thus see (by replacing f with f − λ) that we can reduce the theorem assertion to proving the above inequality in a simpler case when λ = 0.

We thus prove the following inequality:

∫ {x∈ℝ:f∗(x)>0}fdm ≥ 0.

Again, we can use the upwards monotonicity and show instead that for any compact interval [a,b] ⊆R we have

∫ {x∈[a,b]:f∗(x)>0}fdm ≥ 0.

We provide the proof for the simplified assertion above. Fix [a,b]. We apply the rising sun lemma to the continuous function

F : [a,b] → ℝ,F(x) := ∫ [a,x]fdm.

Let’s look at the underlying set. From the rising sun lemma (first paragraph of the proof on the page 118) we know that the set

U := {x ∈ [a,b]∣f∗(x) > 0} = {x ∈ [a,b]∣sup ⁡ h>0∧[x−h,x+h]⊆[a,b] 1 h∫ [x,x+h]fdμ > 0} = {x ∈ [a,b]∣∃ ⁡h > 0 : F(x + h) − F(x) > 0} = {x ∈ [a,b]∣∃ ⁡x < y < b : F(x) < F(y)}

is the union of at most countably many disjoint non-empty open intervals In = (an,bn), with endpoints an,bn lying outside of U. By the first condition of the rising sun lemma, 0 ≤ F(bn) − F(an) = ∫ (an,bn)f, and so we have

∫ {x∈[a,b]:f∗(x)>0}fdm = ∑ n∈ℕ ∫ (an,bn)fdm ≥ 0.

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2021-03-27 00:00
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