Show that $f$ is periodic from an integral property

Question

Let $f$ be a continuous function such that for all $t\in\mathbb{R}$, $$ \int_a^bf(x)\,\mathrm{d}x=\int_{a+t}^{b+t}f(x)\,\mathrm{d}x $$

How can I show that $f$ has to be a periodic function? I was able to prove the statement in the other way, but in this question I end up with $$ \int_a^b(f(x+t)-f(x))\,\mathrm{d}x=0 $$ and from that I don't know what else to do. :(

Answer 1

Define $F(t)=\int_{a+t}^{b+t} f(x) dx = \int_{a}^{b} f(x+t) dx$. You know that $F$ is constant. So $\frac{d}{dt} F(t) = 0$.

On the other hand \begin{align*} \frac{d}{dt} F(t) &= \frac{d}{dt} \int_{a}^{b} f(x+t) dx \newline &= \int_{a}^{b} \frac{d}{dt} f(x+t) dx \newline &= f(b+t)-f(a+t) \newline \end{align*}

Finally $f(b+t)=f(a+t)$ $\forall t\in \mathbb{R}$. You can even use $u=a+t$ so that $f((b-a)+u)=f(u)$ $\forall u\in \mathbb{R}$.