Monthly Archives: June 2015

Pointwise Poincaré inequality

On progress…. I didn’t finish the proof.
We denote \omega_{d} the volume of B\left(x,1\right).

Lemma. Suppose f:\mathbb{R}^{d}\rightarrow\mathbb{R} is continuously differentiable. Then

    \[ \left|f\left(x\right)-\frac{1}{\omega_{d}}\int_{B\left(x,1\right)}f\left(y\right)dy\right|\apprle\int_{B\left(x,1\right)}\frac{\left|\nabla f\left(y\right)\right|}{\left|x-y\right|^{d-1}}dy. \]

Proof. Define F\left(t\right)=f\left(x+t\left(y-x\right)\right). Then F:\left[0,1\right]\rightarrow\mathbb{R} is differentiable and note that F\left(0\right)=f\left(x\right) and F\left(1\right)=f\left(y\right).
So by Fundamental theorem of Calculus, we have F^{\prime}\left(t\right)=\nabla f\left(x+t\left(y-x\right)\right)\cdot\left(y-x\right). So

    \[ f\left(y\right)-f\left(x\right)=F\left(1\right)-F\left(0\right)=\int_{0}^{1}\nabla f\left(x+t\left(y-x\right)\right)\cdot\left(y-x\right)dt. \]

Hence,

    \[ \left|f\left(y\right)-f\left(x\right)\right|\le\left|y-x\right|\int_{0}^{1}\left|\nabla f\left(x+t\left(y-x\right)\right)\right|dt. \]

Then integrate this with respect to y. Then we have

    \begin{align*} \left|f\left(x\right)-\frac{1}{\omega_{d}}\int_{B\left(x,1\right)}f\left(y\right)dy\right| & =\left|\frac{1}{\omega_{d}}\int_{B\left(x,1\right)}\left(f\left(x\right)-f\left(y\right)\right)dy\right|\\  & \apprle\int_{B\left(x,1\right)}\left|x-y\right|\int_{0}^{1}\left|\nabla f\left(x+t\left(y-x\right)\right)\right|dtdy\\  & =\int_{0}^{1}\int_{B\left(x,1\right)}\left|x-y\right|\left|\nabla f\left(x+t\left(y-x\right)\right)\right|dydt. \end{align*}

Take the change of variables z=x+t\left(y-x\right). Then dz=t^{d}dy. So

    \begin{align*}  & \relphantom{=}\int_{0}^{1}\int_{B\left(x,1\right)}\left|x-y\right|\left|\nabla f\left(x+t\left(y-x\right)\right)\right|dydt\\  & =\int_{0}^{1}\int_{B\left(x,t\right)}t^{-1}\left|z-x\right|\left|\nabla f\left(z\right)\right|t^{-d}dzdt\\  & =\int_{B\left(x,1\right)}\left|z-x\right|\left|\nabla f\left(z\right)\right|\int_{\left|z-x\right|}^{1}t^{-\left(d+1\right)}dtdz\\  & \apprle\int_{B\left(x,1\right)}\left|x-z\right|^{-\left(d-1\right)}\left|\nabla f\left(z\right)\right|dz. \end{align*}

As z=x+t\left(y-x\right), take t=1. Then we get

    \[ \left|f\left(x\right)-\frac{1}{\omega_{d}}\int_{B\left(x,1\right)}f\left(y\right)dy\right|\apprle\int_{B\left(x,1\right)}\frac{\left|\nabla f\left(y\right)\right|}{\left|x-y\right|^{d-1}}dy.\qedhere \]

Lebesgue differentiation theorem with approximation to the identity

Lebesgue differentiation theorem은 Riemann integration에서 fundamental theorem of calculus를 measure theory적인 언어로 확장한 정리다. 이를 증명하는 방법이 여러가지가 있으나, 이 절에서는 Stein의 Singular integral and differential properties of functions에 있는 방법을 소개하고자 한다. 이 증명은 약간 good kernel스러운 방법을 쓰고자 한다.

Lebesgue Differentiation Theorem. If f\in L^{1}\left(\mathbb{R}^{d}\right), or more generally if f is locally integrable, then

    \[ \lim_{r\rightarrow0}\frac{1}{m\left(B\left(x,r\right)\right)}\int_{B\left(x,r\right)}f\left(y\right)dy=f\left(x\right), \]

for almost every x.

Proof. Let us denote by f_{r} the function

    \[ f_{r}\left(x\right)=\frac{1}{m\left(B\left(x,r\right)\right)}\int_{B\left(x,r\right)}f\left(y\right)dy,\quad r>0. \]

First, we claim if f\in L^{1}\left(\mathbb{R}^{d}\right), then \lim_{r\rightarrow0}\norm{f_{r}-f}1=0.
Note that for any r>0,

    \begin{align*} f_{r}\left(x\right) & =\frac{1}{m\left(B\left(x,r\right)\right)}\int_{B\left(x,r\right)}f\left(y\right)dy\\ & =\frac{1}{m\left(B\left(x,r\right)\right)}\int_{B\left(x,r\right)}f\left(-y\right)dy\\ & =\frac{1}{\omega_{d}r^{d}}\int_{B\left(0,r\right)}f\left(x-y\right)dy\\ & =\int_{\mathbb{R}^{d}}f\left(x-y\right)\frac{\chi_{B\left(0,r\right)}\left(y\right)}{\omega_{d}r^{d}}dy. \end{align*}

If we define \varphi\left(x\right)=\frac{\chi_{B\left(0,1\right)}\left(x\right)}{\omega_{d}}, then \varphi\ge0 and \int_{\mathbb{R}^{d}}\varphi\left(x\right)dx=1.
Define \varphi_{r}\left(x\right)=r^{-d}\varphi\left(x/r\right). Then \left\{ \varphi_{r}\right\} _{r>0} forms an approximation to the identity.

So

    \begin{align*} f_{r}\left(x\right) & =\int_{\mathbb{R}^{d}}f\left(x-y\right)\varphi_{r}\left(y\right)dy\\ & =\left(f*\varphi_{r}\right)\left(x\right). \end{align*}

This implies \lim_{r\rightarrow0}\norm{f_{r}-f}1=0. Therefore, f_{r_{k}}\rightarrow f, almost everywhere for a suitable sequence \left\{ r_{k}\right\} \rightarrow0. We are left to show that \lim_{r\rightarrow0}f_r\left(x\right) exists almost everywhere.

For this purpose, let us denote for each g\in L^{1}, and x\in\mathbb{R}^{d},

    \[ \omega\left(g\right)\left(x\right)=\left|\limsup_{r\rightarrow0}g_{r}\left(x\right)-\liminf_{r\rightarrow0}g_{r}\left(x\right)\right| \]

where g_{r} is defined like f_{r}. The above quantity represents the oscillation of the family \left\{ g_{r}\right\}, as r\rightarrow0.

Before to prove our theorem, we investigate the property of \omega\left(g\right)\left(x\right) first. If g is continuous with compact support, then g_{r}\rightarrow g uniformly since as we saw before,

    \[ g_{r}\left(x\right)=\left(f*\varphi_{r}\right)\left(x\right) \]

and \left\{ \varphi_{r}\right\} _{r>0} is a family of good kernels since it is an approximation to the identity. Since g is continuous with compact support, g_{r} converges uniformly to g.

If g\in L^{1}\left(\mathbb{R}^{d}\right), then by weak-type inequality, we have

    \[ m\left(\left\{ x:2M\left(g\right)\left(x\right)>\varepsilon\right\} \right)\le\frac{2A}{\varepsilon}\norm g1. \]

However, since g_{r}\left(x\right)\le M\left(g\right)\left(x\right)
for all r>0, we have

    \[ \omega\left(g\right)\left(x\right)\le2M\left(g\right)\left(x\right). \]

Since C_{0}^{\infty}\left(\mathbb{R}^{d}\right) is dense in L^{1}\left(\mathbb{R}^{d}\right), every f\in L^{1}\left(\mathbb{R}^{d}\right) can be written as f=h+g where h is continuous with compact support and g=f-h\in L^{1}\left(\mathbb{R}^{d}\right).

Note that \omega\left(f\right)\le\omega\left(h\right)+\omega\left(g\right) and \omega\left(h\right)=0 since h is continuous with compact support. So

    \[ m\left(\left\{ x:\omega\left(f\right)\left(x\right)>\varepsilon\right\} \right)\le\frac{2A}{\varepsilon}\norm g1. \]

Since the norm of g can be chosen to be arbitrarily small, we get \omega f=0 almost everywhere. So we are done.