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

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.

Maybe you can conclude from the identity that $f$ indeed is differentiable? — mickep, Jun 06 '16 at 04:45
@AyanBiswas yes, use the FTC to show the RHS is differentiable. Then since RHS = LHS you see LHS is as well (because they are the same thing) — Adam Hughes, Jun 06 '16 at 04:47
I think the solution is that $f$ is analytic @Winther and integrating over the unit circle $\int_0^{2 \pi} f'(e^{it}) e^{-i n x} dt$ shows that $f(x) = ax+b$ — reuns, Jun 06 '16 at 05:52
@PedroTamaroff : the $n$th coef in the Taylor series. extracting $c_n z^n$ from $\sum_k c_k z^k$, and using that $c_n z^n$ is solution only if $n = 1$ — reuns, Jun 06 '16 at 05:56
@user1952009 I don't think that is enough based on the example function in an answer in the question linked to above (an analytical function not equal to $a+bz$ which satisfy $f'(z) = f(z+1) - f(z)$). — Winther, Jun 06 '16 at 06:37
@Winther : I did not get where there is a counter-example in your link. — reuns, Jun 06 '16 at 22:07

score 4 · Answer 1 · answered Jun 06 '16 at 12:10

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.

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

1 Answers1