(a) Consider a measure space (X,A,μ) and a complex-valued measurable function F on X. Prove that for any φ:[0,+∞)→[0,+∞) differentiable and increasing such that φ(0)=0, then
∫Xφ(∣F(x)∣)dμ(x)=∫0+∞φ′(s)μ({∣F∣>s})dλ(s)
where λ is the Lebesgue measure.
(b) Consider a complex-valued measurable function f∈L1(Rn)∩L∞(Rn) and its maximal function Mf(x)=supr>0∣B(x,r)∣1∫B(x,r)∣f∣dλ. Prove that for p∈(1,+∞) there is a constant cp>0 such that ∥Mf∥Lp(Rn)⩽cp∥f∥Lp(Rn).
[Hint: Split f=f0+f1 with f0=fχ{∣f∣>s/2} and f1=fχ{∣f∣⩽s/2} and prove that λ({Mf>s})⩽λ({Mf0>s/2}). Then use the maximal inequality λ({Mf>s})⩽ sC1∥f∥L1(Rn) for some constant C1>0.]
(c) Consider p,q∈(1,+∞) with p<q and α∈(0,n) such that 1/q=1/p−α/n. Define Iα∣f∣(x):=∫Rn∣x−y∣n−α∣f(y)∣dλ(y) and prove Iα∣f∣(x)⩽∥f∥Lp(Rn)αp/nMf(x)1−αp/n.
[Hint: Split the integral into ∣x−y∣⩾r and ∣x−y∣∈[2−k−1r,2−kr) for all k⩾0, given some suitable r>0.]