23F (Analysis of Functions)

For every

t \geq 0

, the fundamental theorem of calculus gives

φ (t) = \int_{0}^{t} φ^{'} (s) d s = \int_{0}^{\infty} φ^{'} (s) 1_{{s < t}} d s .

(1)

Apply this with

t = ∣ F (x) ∣

. Since the integrand is nonnegative, Tonelli theorem permits exchanging the integrals:

\int_{X} φ (∣ F (x) ∣) d μ (x) = \int_{X} \int_{0}^{\infty} φ^{'} (s) 1_{{s < ∣ F (x) ∣}} d s d μ (x) = \int_{0}^{\infty} φ^{'} (s) μ ({∣ F ∣ > s}) d s .

(2)

This is the generalized layer-cake representation.

b

 0  0

Solution

 0  0

The layer-cake representation with

φ (t) = t^{p}

gives

∥ M f ∥_{p}^{p} = p \int_{0}^{\infty} s^{p - 1} λ ({M f > s}) d s .

(1)

For each

s > 0

, split

f = f_{0} + f_{1}, f_{0} = f 1_{{∣ f ∣ > s /2}}, f_{1} = f 1_{{∣ f ∣ \leq s /2}} .

(2)

Since

M f_{1} \leq s /2

and the maximal operator is sublinear,

{M f > s} \subseteq {M f_{0} > s /2} .

(3)

The weak

(1, 1)

Hardy-Littlewood maximal inequality therefore gives

λ ({M f > s}) \leq \frac{2 C _{1}}{s} \int_{{∣ f ∣ > s /2}} ∣ f (x) ∣ d x .

(4)

Substitution into the distribution formula and another application of Tonelli theorem yield

∥ M f ∥_{p}^{p} \leq 2 p C_{1} \int_{R^{n}} ∣ f (x) ∣ \int_{0}^{2∣ f (x) ∣} s^{p - 2} d s d x = \frac{2 ^{p} p C _{1}}{p - 1} ∥ f ∥_{p}^{p} .

(5)

Thus the Strong Lp bound for the Hardy-Littlewood maximal function holds with

∥ M f ∥_{p} \leq (\frac{2 ^{p} p C _{1}}{p - 1})^{1/ p} ∥ f ∥_{p} .

(6)

c

 0  0

Solution

 0  0

The relation

1/ q = 1/ p - α / n > 0

implies

α p < n

. Fix

r > 0

and split the Riesz potential into the regions

∣ x - y ∣ < r

and

∣ x - y ∣ \geq r

For the near part, decompose the ball into the dyadic annuli

2^{- k - 1} r \leq ∣ x - y ∣ < 2^{- k} r, k \geq 0.

(1)

The definition of the Hardy-Littlewood maximal function then gives

\int_{∣ x - y ∣ < r} \frac{∣ f ( y ) ∣}{∣ x - y ∣ ^{n - α}} d y \leq C_{n, α} r^{α} M f (x) \sum_{k = 0}^{\infty} 2^{- k α} \leq C r^{α} M f (x) .

(2)

For the far part, Hölder's inequality gives

\int_{∣ x - y ∣ \geq r} \frac{∣ f ( y ) ∣}{∣ x - y ∣ ^{n - α}} d y \leq ∥ f ∥_{p} (\int_{∣ z ∣ \geq r} ∣ z ∣^{- (n - α) p^{'}} d z)^{1/ p^{'}} \leq C ∥ f ∥_{p} r^{α - n / p},

(3)

where convergence follows from

α p < n

M f (x) > 0

, choose

r^{n / p} = \frac{∥ f ∥ _{p}}{M f ( x )} .

(4)

The two bounds then have the same size. If

M f (x) = 0

, the result is immediate. Consequently the Hedberg inequality is

I_{α} ∣ f ∣ (x) \leq C_{n, p, α} ∥ f ∥_{p}^{α p / n} (M f (x))^{1 - α p / n} .

(5)

Here the dimensional constant is required for the unnormalized kernel and Lebesgue measure used in the question.

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Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 23F (Analysis of Functions)

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