TITLE: Every continuous and bounded $f:\mathbb{R}\to\mathbb{R}$ such that $f(x) = \int_x^{x+1} f(y) dy$ for every $x$, is constant
QUESTION [7 upvotes]: Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous and bounded function such that for all $x$:
  $$f(x) = \int_x^{x+1} f(y) dy.$$
  Prove that $f$ is constant.

I could have progressed if it was given that $f$ is differentiable, but with such less info I am finding it tough.

REPLY [4 votes]: Since $f$ is a bounded function it is a tempered distribution. Then the Fourier transformation of
$f'(x) = f(x+1)-f(x)$ gives us
\begin{equation*}
i\xi F(\xi) = e^{i\xi}F(\xi)-F(\xi) \Leftrightarrow F(\xi)\cdot(i\xi -e^{i\xi}+1) = 0
\end{equation*}
where $F$ is the Fourier transform of $f$. But
\begin{equation*}
i\xi -e^{i\xi}+1 = 0 \Leftrightarrow \left\{\begin{array}{l}
\cos \xi = 1\\
\xi -\sin \xi = 0
\end{array}
 \right. \Leftrightarrow \xi = 0 \text{ (double)}.
\end{equation*}
Consequently
\begin{equation*}
\xi^{2}F(\xi) = 0 \Leftrightarrow F(\xi) = A\delta(\xi) + B\delta'(\xi)
\end{equation*}
where $A$ and $B$ are constants. After inverse Fourier transformation we have
\begin{equation*}
f(x) = \dfrac{A}{2\pi} + \dfrac{Bx}{i2\pi}.
\end{equation*}
But $f$ is bounded. Thus $B = 0$ and $f$ is constant.