Paper 1, Section II, F

Analysis of Functions
Part II, 2018

(a) Consider a measure space (X,A,μ)(X, \mathcal{A}, \mu) and a complex-valued measurable function FF on XX. Prove that for any φ:[0,+)[0,+)\varphi:[0,+\infty) \rightarrow[0,+\infty) differentiable and increasing such that φ(0)=0\varphi(0)=0, then

Xφ(F(x))dμ(x)=0+φ(s)μ({F>s})dλ(s)\int_{X} \varphi(|F(x)|) \mathrm{d} \mu(x)=\int_{0}^{+\infty} \varphi^{\prime}(s) \mu(\{|F|>s\}) \mathrm{d} \lambda(s)

where λ\lambda is the Lebesgue measure.

(b) Consider a complex-valued measurable function fL1(Rn)L(Rn)f \in L^{1}\left(\mathbb{R}^{n}\right) \cap L^{\infty}\left(\mathbb{R}^{n}\right) and its maximal function Mf(x)=supr>01B(x,r)B(x,r)fdλM f(x)=\sup _{r>0} \frac{1}{|B(x, r)|} \int_{B(x, r)}|f| \mathrm{d} \lambda. Prove that for p(1,+)p \in(1,+\infty) there is a constant cp>0c_{p}>0 such that MfLp(Rn)cpfLp(Rn)\|M f\|_{L^{p}\left(\mathbb{R}^{n}\right)} \leqslant c_{p}\|f\|_{L^{p}\left(\mathbb{R}^{n}\right)}.

[Hint: Split f=f0+f1f=f_{0}+f_{1} with f0=fχ{f>s/2}f_{0}=f \chi_{\{|f|>s / 2\}} and f1=fχ{fs/2}f_{1}=f \chi_{\{|f| \leqslant s / 2\}} and prove that λ({Mf>s})λ({Mf0>s/2})\lambda(\{M f>s\}) \leqslant \lambda\left(\left\{M f_{0}>s / 2\right\}\right). Then use the maximal inequality λ({Mf>s})\lambda(\{M f>s\}) \leqslant C1sfL1(Rn)\frac{C_{1}}{s}\|f\|_{L^{1}\left(\mathbb{R}^{n}\right)} for some constant C1>0.]\left.C_{1}>0 .\right]

(c) Consider p,q(1,+)p, q \in(1,+\infty) with p<qp<q and α(0,n)\alpha \in(0, n) such that 1/q=1/pα/n1 / q=1 / p-\alpha / n. Define Iαf(x):=Rnf(y)xynαdλ(y)I_{\alpha}|f|(x):=\int_{\mathbb{R}^{n}} \frac{|f(y)|}{|x-y|^{n-\alpha}} \mathrm{d} \lambda(y) and prove Iαf(x)fLp(Rn)αp/nMf(x)1αp/nI_{\alpha}|f|(x) \leqslant\|f\|_{L^{p}\left(\mathbb{R}^{n}\right)}^{\alpha p / n} M f(x)^{1-\alpha p / n}.

[Hint: Split the integral into xyr|x-y| \geqslant r and xy[2k1r,2kr)|x-y| \in\left[2^{-k-1} r, 2^{-k} r\right) for all k0k \geqslant 0, given some suitable r>0.]r>0 .]