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.)

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.
