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 $f(x+\theta)-f(x)\ne0$ 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.)

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}$.



*

*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 $f(0) = f(1)$.

*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.

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.

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.
