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Spring 2015 - Problem 3

Hardy-Littlewood maximal inequality

Let f∈Lloc1(Rn)f \in L^1_{\mathrm{loc}}\p{\R^n} and let

Mf(x)=sup⁡r>01m(B(r,x))∫B(r,x)∣f(y)∣ dyMf\p{x} = \sup_{r > 0} \frac{1}{m\p{B\p{r, x}}} \int_{B\p{r,x}} \abs{f\p{y}} \,\diff{y}

be the Hardy-Littlewood maximal function.

  1. Show that

    m({x∣Mf(x)>s})≤Cns∫{∣f(x)∣>s/2}∣f(x)∣ dx,s>0,m\p{\set{x \mid Mf\p{x} > s}} \leq \frac{C_n}{s} \int_{\set{\abs{f\p{x}} > s/2}} \abs{f\p{x}} \,\diff{x}, \quad s > 0,

    where the constant CnC_n depends on nn only. The Hardy-Littlewood maximal theorem may be used.

  2. Prove that if φ∈C1(R)\phi \in C^1\p{\R}, φ(0)=0\phi\p{0} = 0, and φ′>0\phi' > 0, then

    ∫φ(Mf(x)) dx≤Cn∫∣f(x)∣(∫{0<t<2∣f(x)∣}φ′(t)t dt) dx.\int \phi\p{Mf\p{x}} \,\diff{x} \leq C_n \int \abs{f\p{x}} \p{\int_{\set{0 < t < 2\abs{f\p{x}}}} \frac{\phi'\p{t}}{t} \,\diff{t}} \,\diff{x}.
Solution.

  1. Let g=fχ{∣f(x)∣>s/2}g = f\chi_{\set{\abs{f\p{x}} > s/2}} and h=fχ{∣f(x)∣≤s/2}h = f\chi_{\set{\abs{f\p{x}} \leq s/2}} so that f=g+hf = g + h. Observe that for all x∈Rnx \in \R^n and r>0r > 0, we have

    1m(B(x,r))∫B(x,r)∣h(y)∣ dy≤s2  ⟹  Mh(x)≤s2.\frac{1}{m\p{B\p{x,r}}} \int_{B\p{x,r}} \abs{h\p{y}} \,\diff{y} \leq \frac{s}{2} \implies Mh\p{x} \leq \frac{s}{2}.

    Thus, {x∣Mh(x)>s2}=∅\set{x \mid Mh\p{x} > \frac{s}{2}} = \emptyset, and so {x∣Mf(x)>s}⊆{x∣Mg(x)>s2}\set{x \mid Mf\p{x} > s} \subseteq \set{x \mid Mg\p{x} > \frac{s}{2}}. Hence,

    m({x∣Mf(x)>s})≤m({x∣Mg(x)>s2})≤Cn′s/2∥g∥L1(Hardy-Littlewood maximal inequality)=Cns∫{∣f(x)∣>s/2}∣f(x)∣ dx.\begin{aligned} m\p{\set{x \mid Mf\p{x} > s}} &\leq m\p{\set{x \mid Mg\p{x} > \frac{s}{2}}} \\ &\leq \frac{C_n'}{s/2} \norm{g}_{L^1} && \p{\text{Hardy-Littlewood maximal inequality}} \\ &= \frac{C_n}{s} \int_{\set{\abs{f\p{x}} > s/2}} \abs{f\p{x}} \,\diff{x}. \end{aligned}

  2. By the fundamental theorem of calculus and the fact that φ(0)=0\phi\p{0} = 0,

    ∫φ(Mf(x)) dx=∫Rn∫0Mf(x)φ′(t) dt dx=∫Rn∫0∞χ{0<t<Mf(x)}φ′(t) dt dx=∫0∞∫Rnχ{0<t<Mf(x)}φ′(t) dx dt(Fubini-Tonelli)=∫0∞m({0<t<Mf(x)})φ′(t) dt≤∫0∞φ′(t)Cnt∫{∣f(x)∣>t/2}∣f(x)∣ dx dt(1)=Cn∫φ′(t)t∫χ{0<t<2∣f(x)∣}∣f(x)∣ dx dt=Cn∫∣f(x)∣(∫{0<t<2∣f(x)∣}φ′(t)t dt) dx,(Fubini-Tonelli)\begin{aligned} \int \phi\p{Mf\p{x}} \,\diff{x} &= \int_{\R^n} \int_0^{Mf\p{x}} \phi'\p{t} \,\diff{t} \,\diff{x} \\ &= \int_{\R^n} \int_0^\infty \chi_{\set{0 < t < Mf\p{x}}} \phi'\p{t} \,\diff{t} \,\diff{x} \\ &= \int_0^\infty \int_{\R^n} \chi_{\set{0 < t < Mf\p{x}}} \phi'\p{t} \,\diff{x} \,\diff{t} && \p{\text{Fubini-Tonelli}} \\ &= \int_0^\infty m\p{\set{0 < t < Mf\p{x}}} \phi'\p{t} \,\diff{t} \\ &\leq \int_0^\infty \phi'\p{t} \frac{C_n}{t} \int_{\set{\abs{f\p{x}} > t/2}} \abs{f\p{x}} \,\diff{x} \,\diff{t} && \p{\text{1}} \\ &= C_n \int \frac{\phi'\p{t}}{t} \int \chi_{\set{0 < t < 2\abs{f\p{x}}}} \abs{f\p{x}} \,\diff{x} \,\diff{t} \\ &= C_n \int \abs{f\p{x}} \p{\int_{\set{0 < t < 2\abs{f\p{x}}}} \frac{\phi'\p{t}}{t} \,\diff{t}} \,\diff{x}, && \p{\text{Fubini-Tonelli}} \end{aligned}

    which completes the proof.