Continuous $f:[0,1]\to\mathbb{R}$ such that $f(0)=f(1)$ and $\forall\alpha\in(0,1)\exists c\in[0,1-\alpha]|f(c)=f(c+\alpha)$?
Solution 1:
If you choose $$f(x):= \begin{cases} x &: x < \frac{1}{4} \\ \frac{1}{2}-x &: \frac{1}{4}\leqslant x\leqslant \frac{3}{4} \\ x-1 &: \frac{3}{4}\leqslant x \leqslant 1\end{cases}$$ and $\alpha=\frac{3}{4}$ then $f(x)\geq 0$ and $f(x+\alpha)\leqslant 0$ for all $x\in [0,1-\alpha]$.
Solution 2:
Consider $f(x) = \sin(2\pi x)$ and $1/2 < \alpha < 1$. Then, $f(x)\geq 0$ and $f(x+\alpha)\leq 0$ for all $x\in [0, 1-\alpha]$.