For $0\lt\theta\lt1$, $\frac 1\theta\notin\mathbb Z$, there exists $f\in C[0, 1]$ such that $f(0)=f(1)$ and $f(x+\theta)-f(x)\ne0$

Question

Prove that for each $0\lt\theta\lt1, \dfrac{1}{\theta} $ isn't an integer, there exists $f \in C[0, 1]$ such that $f(0)=f(1)$, and $ \forall x\in[0,1-\theta] , f(x+\theta)-f(x)\ne0 $

(If $\theta\gt\frac12, $ it is obvious that such an $f$ exists.)

Answer 1

Since I cannot close as dupe because of the bounty.

Here is prior art: Universal Chord Theorem

Answer 2

Parting from predicates, as $0 < \theta < 1$ in strict order, we can establish a function $\phi : ]0,1[ \rightarrow \mathbb{R}$ such that $\phi(\theta) = \frac{1}{\theta}$. As we know that $\phi(\theta)$ tends to $1$ as $\theta$ tends to $1$, and $+ \infty$ as $\theta$ tends to $0$, we can restrict the function as follows:

$$ \phi : ]0,1[ \rightarrow ]1,+\infty[ \\ \theta \mapsto \frac{1}{\theta} $$

Note that as $\mathbb{R}$ is a dense set, $]0,1[$ is homeomorphic to $ ]1,+\infty [ \subset \mathbb{R}$.

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1. If such an $f$ existed for all such $\theta$, we'd only need continuity to conclude by Rolle's Theorem that, a subset of all these functions is indeed defined such that $f(0) = f(1)$.
2. Also, if such an $f$ existed for all such $\theta$, we'd have:

$$ \forall x \in [0,1-\theta], \exists a \ne 0 : f(x+\theta) - f(x) = a $$

then, as $\theta \ne 0$,

$$ \frac{f(x+\theta) - f(x)}{x+\theta-x} = f’(\theta) = \frac{a}{\theta} \ne 0 $$

Thus, $f$ is differentiable on $[0,1 - \theta]$ and non-constant.

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Let $\alpha \in ]1,+\infty[$,

Then there exists $\theta \in ]0,1[$ such that $\phi (\theta) = \frac{1}{\theta} = \alpha$.

Indeed, we can consider the function:

$$ f(x) = sin(2 \alpha \pi x) $$

The function $f$ defined above -luckily- holds for an example for each choice of $\theta \in ]0,1[$ such that $\alpha = \phi^{-1}(\theta)$ whose existence is guaranteed by construction. That is why I won't discuss the property of $\frac{1}{\theta}$ being non-integer. It is a continuous function and clearly not constant over $[0,1-\theta]$, which we will extend by continuity to its values on ${0}$ and ${1}$ too.

We then have the existence for them all.

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Furthermore, the only fact that any step interval $ ]n,n+1[ $ with $n \in \mathbb{N}$ is homeomorphic to $ ]1,+\infty[ $ leads to the existence of a family $F$ of such functions.

Notice that, we can generalize the subset of $F$ generated by these $f$ above:

$$ \forall n \in \mathbb{N}, f_{n}(x) = sin(2 (n+1) \alpha \pi x) $$

Observe that each element of $ \{ f_{i} \}_{i \in I_n} $ is in $F$.

Thus the generalized form.

Answer 3

If $\frac12<\theta<1$, take $f(x)=\sin(2\pi x)$.

If $0<\theta<\frac12$, since $\frac1\theta\not\in\mathbb Z$, there exists a positive integer $m\ge2$ such that $m\theta<1<(m+1)\theta$. Let \begin{cases} a_k=k\theta,\\ b_k=1-(m-k)\theta \end{cases} for $k=0,1,2,\ldots$. Then $a_0=0,\ b_m=1$ and $a_0<b_0<a_1<b_1<\cdots$. Now, let $f:[0,\infty)\to\mathbb R$ be the polygonal line joining the nodes $$ \left(a_0,f(a_0)\right),\,\left(b_0,f(b_0)\right),\,\left(a_1,f(a_1)\right),\,\left(b_1,f(b_1)\right),\ldots $$ where \begin{cases} f(a_k)=k+1,\\ f(b_k)=k-m \end{cases} for each $k$. By construction, $f$ is continuous and $f(0)=f(a_0)=0=f(b_m)=f(1)$. Also, for every $k$, we have \begin{cases} f(a_k+\theta)=f(a_{k+1})=k+1=f(a_k)+1,\\ f(b_k+\theta)=f(b_{k+1})=k+1-m=(k-m)+1=f(b_k)+1. \end{cases} Therefore $f(x+\theta)=f(x)+1$ at every node (with $x\in\{a_0,b_0,a_1,b_1,\ldots\}$) on the polygonal line. Consequently, by linearity of $f$ between consecutive nodes, the equality $f(x+\theta)=f(x)+1$ must hold for every $x$. Hence our $f$ is a feasible choice.