3

Prove that if $E \subset [0,1]$ has positive measure, then the set $E-E = \{x-y : x,y \in E\}$ contains an interval centered around zero. Hint: consider the function $h(x)=\textbf{1}_{-E} \star \textbf{1}_{E} $.

Ideas: use the continuity of h(x)?

user62108
  • 181
  • Why would $h$ be continuous? – Frank Mar 25 '14 at 22:15
  • Yes, using the continuity of $h$ is a good idea. You need to say why it is continuous, though. – Daniel Fischer Mar 25 '14 at 22:18
  • I know why h(x) is continuous but I'd rather not write it out on here. It was part of a previous exercise. Essentially if $f\in L^1$ and $g \in L^{\infty}$ then $f \star g$ is continuous. So my idea is to take the pre-image of some set to get an interval around zero. – user62108 Mar 25 '14 at 22:59
  • Just saying that $h$ is continuous because both functions belong to $L^1\cap L^\infty$ is sufficient then. Now to reach the conclusion, find an $x$ such that $h(x) > 0$. – Daniel Fischer Mar 25 '14 at 23:04

1 Answers1

4

$$h(x)= \int 1_E(y)1_{-E}(x-y)dy= \int 1_E(y)1_{E(x+E)}(y)dy=m(E\cap (x+E))$$ Prove that $h$ is continuous. Then observe that $h(0)=m(E)>0$, so $0\in h^{-1}(0,\infty)$ which is an open set. Therefore, there exists an open interval around $0$, say $(-\varepsilon,\varepsilon)$ contained in $h^{-1}(0,\infty)$. Show that this is the desired interval.

user10354138
  • 33,887
Dimitris
  • 7,190
  • 1
  • 32
  • 49
  • Why is this function continuous? – Ángela Flores May 03 '18 at 05:02
  • In general if you take the convolution of a bounded function with an $L^1$ function you get a continuous function. See https://math.stackexchange.com/questions/570494/convolution-is-uniformly-continuous-and-bounded – Dimitris May 04 '18 at 04:32
  • Spoiler alert: to show that this is the desired interval: take any $p\in (-\epsilon,\epsilon)$. Then $0<h(p)=m(E\cap p+E)$ so $E\cap p+E \not = \emptyset$ hence there are $e_1,e_2 \in E$ with $e_1=p+e_2$... – Marcelo Nov 18 '18 at 00:37