Steinhaus Theorem in $\mathbb{R}^d$

Question

I'm having some trouble proving the Steinhaus theorem in $\mathbb{R}^d$:

Claim: Let $E\subset\mathbb{R}^d$ be a (measurable) set with positive measure. For some $\epsilon\gt0$, $E-E=\{x-y:x,y\in E\}$ contains the open ball $B_{\epsilon}(0)$ of radius $\epsilon$ centered at 0.

We were given the hint to use the fact that $f*g$ with $f(x)=1_{E}(x)$ and $g(x)=1_E(-x)$ is continuous. But I have no idea, how to use this. Any help would be greatly appreciated.

score 10 · Answer 1 · edited May 26 '19 at 14:25

10

Given that you know that the function $$ \varphi(x) = \int_{\mathbb{R}^n} \chi_E(y)\chi_E(x+y)dy $$ is continuous. Note that $$ \varphi(x) = \int_E \chi_E(x+y)dy = \int_{E+x}\chi_E(y)dy = \mu(E\cap (E+x)) $$ Hence, $\varphi(0) = \mu(E) > 0$, so there is a neighbourhood of $0$ on which $\varphi$ must be positive.

Now check that $\varphi(x) > 0$ implies that $x\in E-E$

edited May 26 '19 at 14:25

Bach

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answered Dec 19 '13 at 16:31

Prahlad Vaidyanathan

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Thanks for your help! Just one question: Shouldn't $\varphi(x)$ be the integral of $\chi_{E}(y)\chi_{E}(x+y)$? – caedmon Dec 19 '13 at 17:47
caedmon I think you are correct.
This is the slickest proof I have seen of this fact. The only way I knew was in $d = 1$; one can start by showing that for any $\alpha \in (0,1)$ there is a bounded interval $I$ such that $m(A \cap I) \geq \alpha m(I)$. Can it really be this simple?
– snar Dec 19 '13 at 17:58
@snarski : It actually takes a fair amount of work to prove that $\varphi$ is continuous (try it!). – Prahlad Vaidyanathan Dec 20 '13 at 03:37
@PrahladVaidyanathan I don't see it being any different than showing that the convolution $f*g$ of two $L^1$ functions being continuous, which as you say is not immediate, but at least I know how to do. Is that right? – snar Dec 20 '13 at 03:51
@snarski : yes, but in that case, you need to assume $\mu(E) < \infty$. – Prahlad Vaidyanathan Dec 20 '13 at 03:55
@PrahladVaidyanathan Ah, right. So do it for finite sets and then use monotone convergence theorem on $E_n := E \cap B_n(0)$? – snar Dec 20 '13 at 03:59
The convolution of two $L^1$ functions is not continuous in general. But the convolution of an $L^1$ and an $L^\infty$ function is, which is enough here. – PhoemueX May 01 '20 at 06:18

score 0 · Answer 2 · edited Apr 13 '17 at 12:19

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The method I present in this answer easily generalizes to $\mathbb R^d$. It is outlined in exercise 5 of chapter 7 (differentiation) in Rudin's Real and Complex Analysis.

edited Apr 13 '17 at 12:19

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answered Jun 30 '15 at 16:07

Potato

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Steinhaus Theorem in $\mathbb{R}^d$

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