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This is a question on a proof in the paper "Baire Category, Probabilistic Constructions and Convolution Squares" by TW Körner (Theorem 4.10 page 13). The theorem states that in the complete metric space $\mathcal{G}= \{ (E_1,E_2): E_1,E_2 \text{ are compact subsets of} \ \mathbb{T}, E_1+E_2 =\mathbb{T} \}$ with metric $d((E_1,E_2),(F_1,F_2))= \Delta(E_1,F_1) + \Delta(E_2,F_2)$ ($\Delta$ denotes the Hausdorff distance), the set of elements $(E_1,E_2)$ that are pairs of Kronecker sets is dense and open in $\mathcal{G}$.

Here $S(\mathbb{T})=\{ f \in C(\mathbb{T}): |f|=1 \}$ and $\{ f_j \} $ a countable dense subset of $S(\mathbb{T})$.

The statement I don't understand is this:

"By the uniform continuity of $f_j$ and the intermediate value theorem, any sufficiently large M will have the property that the equation $f_j(t)=e^{2\pi i M t}$ will have at least one solution in any closed interval $I$ of length $\frac{\epsilon}{ 4}$."

If one chooses an irrational number $\xi \in (0,1)$ the sequence $\{ e^{2\pi i n \xi} \}_{n\geq 1} $ will be dense in $\mathbb{T}$ by Weyl's equidistribution theorem. Does it have something to do with this?

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  • $\begingroup$ Could one use some sort of fixed point theorem on $f_j(t) - e^{2\pi i M t}$? $\endgroup$ – user202542 Jan 8 '18 at 13:12
  • $\begingroup$ I was looking around on the forum and found this version of the intermediate value theorem: math.stackexchange.com/questions/1579953/…. I think that one can use that if one can show that for an interval/arc $I \subset \mathbb{T}$ and for a large enough $M$ $f_j(I) \supset (I)^M$. Is this true? $\endgroup$ – user202542 Jan 9 '18 at 16:32
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The intuition is quite simple: By selecting $M$ really large, over any interval of width $\frac{\epsilon}{4}$ the map $t \mapsto e^{2\pi iMt}$ will complete as many revolutions of the circle as you like; but on the other hand, there are only so many complete revolutions of the circle (if any) that a given continuous function $f_j$ can complete over an interval of width $\frac{\epsilon}{4}$. Hence, by selecting $M$ large enough, you can guarantee that on any interval of width $\frac{\epsilon}{4}$ the two functions will hit each other (in fact, as many times as you like).

But now let's make this more rigorous:

For any $t \in \mathbb{R}$, write $[t] \in \mathbb{T} \simeq \mathbb{R}/\mathbb{Z}\ $ for the projection of $t$.

For all $t$, let $\, D([t]):=F(t+\frac{\epsilon}{4})-F(t)$,$\,$ where $F(\cdot)$ is a continuous function satisfying $\,f_j([t])=e^{2\pi iF(t)}$.$\,$ Note that $D(\cdot)$ is a continuous function on $\mathbb{T}$, and so since $\mathbb{T}$ is compact, $D(\cdot)$ is bounded.

So let $K<\infty$ be such that $D(\cdot)<K$ on the whole of $\mathbb{T}$.$\,$ Take $M$ sufficiently large that $M.\!\frac{\epsilon}{4}>K+1$.

Now let $\,I \subset \mathbb{T}\;$ be any arc of length $\frac{\epsilon}{4}$, and let $u \in \mathbb{R}$ be such that $[u]$ and $[u+\frac{\epsilon}{4}]$ are respectively the left and right endpoints of $I$. Fixing a function $F(\cdot)$ as above, let $n \in \mathbb{Z}$ be such that $Mu-n \in [F(u)-1,F(u))$. Then $$ M(u+\tfrac{\epsilon}{4})-n \,\geq\, F(u)-1+M.\!\tfrac{\epsilon}{4} \,>\, F(u)+K \,>\, F(u)+D([u]) \,=\, F(u+\tfrac{\epsilon}{4}). $$ So then, $\ Mu-n<F(u)\ $ while $\ M(u+\frac{\epsilon}{4})-n>F(u+\frac{\epsilon}{4})$.$\,$ Hence, by the "intermediate value theorem for two functions" (as linked in your comment), there exists $t \in (u,u+\frac{\epsilon}{4})$ such that $Mt - n = F(t)$ and therefore $f_j([t])=e^{2\pi i Mt}$.$\,$ Since $t \in (u,u+\frac{\epsilon}{4})$, we have that $[t] \in I$.

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  • $\begingroup$ The boundedness of the function $D(\cdot)$ can also be seen as a consequence of the uniform continuity of $f_j$, as is implicit within the quote you gave; but that is a rather round-about approach. $\endgroup$ – Julian Newman Jan 10 '18 at 2:23

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