Pointwise Poincaré inequality

By | June 11, 2015

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

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

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