Let $f:[0, 1]\rightarrow \mathbb{R}$ a continuous function such that $f(0)=f (1)=0$ and we define
$A=${$h\in[0, 1]|$ there exists $x$ with $f(x+h)=f(x)$}
$B=${$h\in[0, 1]|$ there exists $x$ with $f(x+1-h)=f(x)$}
Prove that $A\cup B=[0, 1]$.
I tried using the fact that $f$ is continuous but since $f$ has values in $\mathbb{R}$ it is a little bit more complicated. I assumed that there exists an $h\in[0, 1]$ sucht that none of the relations was true but I didn't managed to show anything.
It is easy to see that $0,1\in A\cap B$. Let $h\in (0,1)$ and consider the continuous function $$g(x):=\begin{cases} f(x)&\text{if $x\in [0,1]$,}\\ f(x-1)&\text{if $x\in [1,2]$.} \end{cases}$$ then, since $$\int_0^1 (g(x + h) - g(x)) dx=\int_h^{1+h} g(x)dx - \int_0^1g(x) dx=0,$$ it follows that there exists $x\in [0,1)$ such that $g(x+h)-g(x)=0$.
Now, if $x+h\leq 1$ then $f(x+h)=f(x)$ and $h\in A$.
On the other hand, if $x+h>1$ then $f(y)=f(y+1-h)$ for $y:=x+h-1\in(0,1)$ and $h\in B$.
Therefore we may conclude that $h\in A\cup B$.