# Find $f$ such that $\{f(x+y)\}=\{f(x)\}+\{f(y)\}$

Find all continuous function such that $\{f(x+y)\}=\{f(x)\}+\{f(y)\}$ for all $x, y\in\mathbb{R}$. Denote $\{x\}=x-[x]$ in which $[x]$ is the largest integer number does not exceed x.

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If $f$ ever crosses an integer boundary, say at $x$, there's a jump by $\pm1$ in $\{f(x)\}$. Then for given $y$, there would have to be a corresponding jump in $\{f(x+y)\}$. But since we can choose $y$ arbitrarily, that would mean that $f$ would have to cross an integer boundary at every point, which is impossible.

Thus $f$ never crosses an integer boundary. But then $g(x)=\{f(x)\}$ is a continuous function with $g(x+y)=g(x)+g(y)$. An additive continuous function must be linear; but the only linear function that can be the fractional part of $f$ is the zero function. Thus $f$ takes a constant integer value.

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Just a minor nit: $f(x)$ is a constant. Nothing says it has to be an integer. –  vonbrand Feb 3 '13 at 5:19
@vonbrand: Why not? $g$ is linear; since it's a fractional part, it's constrained to $[0,1)$; so it must be the zero function. If the fractional part of $f$ is zero, $f$ is an integer. –  joriki Feb 3 '13 at 6:01
Too little coffee. Thanks. –  vonbrand Feb 3 '13 at 6:05
For all $x \in \mathbb{R}$, we have $0 \leq \{x\} <1$.
Suppose $\{f(x+y)\} = \{f(x)\} + \{f(y)\}$. Can you show that there cannot be any $x$ such that $\{f(x)\} \geq 0.5$?
Can you show that there cannot be any $x$ such that $\{f(x)\}>0$?