1.6. Differentiation theorems 119 Now we can prove the one-sided Hardy-Littlewood maximal inequality. By upwards monotonicity, it will suffice to show that m({x [a, b] : sup h0 [x,x+h]⊂[a,b] 1 h [x,x+h] |f(t)| dt λ}) 1 λ R |f(t)| dt for any compact interval [a, b]. By modifying λ by an epsilon, we may replace the non-strict inequality here with strict inequality: (1.25) m({x [a, b] : sup h0 [x,x+h]⊂[a,b] 1 h [x,x+h] |f(t)| dt λ}) 1 λ R |f(t)| dt. Fix [a, b]. We apply the rising sun lemma to the function F : [a, b] R defined as F (x) := [a,x] |f(t)| dt (x a)λ. By Corollary 1.6.5, F is continuous, and so we can find an at most countable sequence of intervals In = (an,bn) with the properties given by the rising sun lemma. From the second property of that lemma, we observe that {x [a, b] : sup h0 [x,x+h]⊂[a,b] 1 h [x,x+h] |f(t)| dt λ} n In, since the property 1 h [x,x+h] |f(t)| dt λ can be rearranged as F (x + h) F (x). By countable additivity, we may thus upper bound the left-hand side of (1.25) by n (bn an). On the other hand, since F (bn) F (an) 0, we have In |f(t)| dt λ(bn an) and thus n (bn an) 1 λ n In |f(t)| dt. As the In are disjoint intervals in I, we may apply monotone convergence and monotonicity to conclude that n In |f(t)| dt [a,b] |f(t)| dt, and the claim follows. Exercise 1.6.11 (Two-sided Hardy-Littlewood maximal inequality). Let f : R C be an absolutely integrable function, and let λ 0. Show that m({x R : sup x∈I 1 |I| I |f(t)| dt λ}) 2 λ R |f(t)| dt, where the supremum ranges over all intervals I of positive length that con- tain x.
Previous Page Next Page