Does the Compact Set K+K Have Nonempty Interior?

[Nedeljko]

Let $$K\subseteq\mathbb R$$ is a compact set of positive Lebesque measure. Prove that the set $$K+K=\{a+b\,|\,a,b\in K\}$$ has nonempty interior.

 

[morphism]

What have you tried so far? Do you know of any links between the Lebesgue measure and the topology of R?

 

[Nedeljko]

All open and all closed sets are measurable. For any measurable set $$E$$ and positive real $$\varepsilon$$ there are an closed set $$F\subseteq E$$ and an open set $$G\supseteq E$$ such that measure of $$G\setminus F$$ is less than $$\varepsilon$$.

 

[Nedeljko]

The original problem is formulated for any measurable set $$E$$. Using the fact that any measurable set of positive measure has a compact subset of positive measure, the general case is reduced to special case of compact sets.

 

[morphism]

I thought I had an elementary solution, but I discovered some holes in my reasoning. Anyway, a quick and dirty way to solve the problem is to convolve the characteristic function of K with itself. You will get a nonnegative continuous function f whose integral is m(K)^2 > 0. Thus f must be positive on some open interval - but f is positive precisely on K+K.

 

[Nedeljko]

Are you sure that convolution is continuous? Why? Do you have a link? You state that the convolution is positive exactly on K+K, but K+K is compact and continuous function is positive on the open set.

 

[morphism]

It's a theorem that if $f \in L^1(\mathbb{R})$ and $g \in L^\infty(\mathbb{R})$ then $f*g$ is continuous. (In fact, the same conclusion holds for $f \in L^p(\mathbb{R})$ and $g \in L^q(\mathbb{R})$, where p and q are convex conjugates.) You can find this in almost any text on harmonic analysis, e.g. it's on page 4 of Rudin's Fourier Analysis on Groups.

The point of my post was the following: Since f is positive precisely on K+K (this statement actually requires a line of proof) and since f is positive on an open interval, then K+K must contain an open interval.

 

[gel]

that's very neat.
One thing though, f is positive on a subset of K+K (rather than precisely on K+K), which is exactly what was needed - and the condition that K is compact wasn't needed, only finite measure.

...and to show that convolution f*g is continuous is easy when g is continuous. The general case follows by taking limits.

 

[Nedeljko]

I proved thjat $$f$$ is continuous an that the open set $$U=\{x\in\mathbb{R}\,:\,f(x)>0\}$$ is the subset of $$K+K$$. But, how to prove that the set $$U$$ is nonempty? In fact, $$f(x)=m(K\cap(x-K))$$.

 

[gel]

Note that if {f>0} was empty then the integral of f would be zero. Can you say what the integral of f is?

 

[Nedeljko]

OK, application of Fubini's theorem gives $$\int_{\mathbb{R}}f\,dm=m(K)^2>0$$. Thanks.