$f(1) = 1$, $f(x+5)\geq f(x)+5$, and $f(x+1)\leq f(x) +1$ for any $x\in\mathbb{R}$

Question

Let $f(1) = 1$, $f(x+5)\geq f(x)+5$, and $f(x+1)\leq f(x) +1$ for any $x\in\mathbb{R}$. I have shown that the following statements are true :

1. $f(x+n) = f(x) + n$ for any $x\in\mathbb{R}$ and for any $n\in\mathbb{Z}$,
2. $f(x) = x$ for any $x\in\mathbb{Z}$.

My question : Is it possible to prove that the first statement holds when $n\in\mathbb Z$ is replaced by $n\in\mathbb{R}\setminus\mathbb{Z}$? Do we need additional information? (For instance, if $f$ is continuous then it is obvious)

Answer 1

No, we can not replace $\Bbb Z$ with $\Bbb R\setminus\Bbb Z$.

Suppose $g(x)$ is an arbitrary function from interval $(0,1)$ to $\Bbb R$. Let $f_g: \Bbb R\to\Bbb R$, $f_g(x)=x$ if $x\in\Bbb Z$, $f_g(x)=\lfloor x\rfloor + g(\{x\})$ if $x\notin\Bbb Z$, where $\lfloor x\rfloor$ and $\{x\}$ are the integral part and the fractional part of $x$. Verify that $f_g(x)$ satisfies the conditions.

To make $f_g(x)$ continuous, just let $g(x)$ be continuous, $\lim_{x\to 0^+}g(x)=0$ and $\lim_{x\to 1^-}g(x)=1$.

However, this additional information does not help ensure $f_g(x+n)=f_g(x)+n$ for $n\notin\Bbb Z$ much. For example, if $g(x)=x^2$, then $f_g(x)$ is continuous. But $$f_g(\frac34+\frac14)=1\not=\frac9{16}+\frac14=f_g(\frac34)+\frac14.$$

The information needed is $f(x)=x$ on $(0,1)$.