My Math Forum Proving an Inequality for Locally Integrable Functions

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 April 7th, 2019, 01:52 PM #1 Newbie   Joined: Apr 2019 From: Madison, Wisconsin Posts: 2 Thanks: 0 Math Focus: Measure Theory & Integration Proving an Inequality for Locally Integrable Functions >$\textbf{The Problem:}$ Let $f\geq 0$ be a bounded function and $E\subset\mathbb R^d$ have finite measure. Prove that there exists $R>0$ such that for all $00$ does not exist. Then given $R>0$ there is some $0\int_{E}f(x)dx> 2\int_{E}\left(\frac{1}{\vert B(x,r)\vert}\int_{B(x,r)}f(y)dy\right)dx.$$In particular, this means that for almost every x\in E we must have$$\infty>\frac{f(x)}{2}>\frac{1}{\vert B(x,r)\vert}\int_{B(x,r)}f(y)dy.$$And since by the assumption that f\geq0 and bounded, we have that$$\frac{1}{\vert B(x,r)\vert}\int_{B(x,r)}f(y)dy\geq \frac{1}{\vert B(x,p)\vert}\int_{B(x,p)}f(y)dy$$for all 0 April 7th, 2019, 04:40 PM #2 Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics The following line is wrong. Quote:  Originally Posted by Gaby Alfonso Then given R>0 there is some 0\int_{E}f(x)dx> 2\int_{E}\left(\frac{1}{\vert B(x,r)\vert}\int_{B(x,r)}f(y)dy\right)dx.$$ In particular, this means that for almost every$x\in E\$ we must have $$\infty>\frac{f(x)}{2}>\frac{1}{\vert B(x,r)\vert}\int_{B(x,r)}f(y)dy.$$
This is not true. You have taken an inequality for integrals and tried to use it to imply a pointwise inequality which in general is not valid even on a set of full measure.

Last edited by skipjack; April 8th, 2019 at 12:20 AM.

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