If $f(1)=c$ and $f(x+1) = f(x) + f(1)$, then is $f(x + y) = f(x) + f(y)$?

Asked Aug 14 '22 at 01:34

Active Aug 14 '22 at 02:10

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Suppose that $f: \mathbb{R} \to \mathbb{R}$ and let $c \in \mathbb{R} \backslash \{0\}$ be fixed such that:

I proved the following properties:

For any $m \in \mathbb{Z}$ and any $x \in \mathbb{R}$, $$f(x + m) = f(x) + f(m) = f(x) + cm$$
$f$ is well-defined.

It seems that $f$ satisfy:

For all $x,y \in \mathbb{R}$, $f(x+y) = f(x) + f(y)$

but I dont think its true, but if I can show that property, maybe $f(x) = cx$, for all $x \in \mathbb{R}$.

Maybe using $x = \lfloor x \rfloor - \text{frac}(x)$ will work, but it got me with this property:

For any $x \in \mathbb{R}$: $$f(x) - f(\text{frac}(x)) = f(x - \text{frac}(x))$$

Any hints?

Thanks.

EDIT

I noticed that $f$ is not injective. My bad.

edited Aug 14 '22 at 02:10

asked Aug 14 '22 at 01:34

Joshua Ortiz

3

Consider $f(x)=\sin(2\pi x)+cx$. – John Wayland Bales Aug 14 '22 at 02:01
Thank you, @JohnWaylandBales! – Joshua Ortiz Aug 14 '22 at 02:05
@JohnWaylandBales, is that the unique solution? – Joshua Ortiz Aug 14 '22 at 02:25
2

No, let $f(x)=g(x)+cx$ where $g$ is periodic of period $1$. – John Wayland Bales Aug 14 '22 at 02:41
I forgot to add "and $g(0)=0$." – John Wayland Bales Aug 14 '22 at 18:57
@JohnWaylandBales, why not "$g(m) = 0$, for all $m \in \mathbb{Z}$? – Joshua Ortiz Aug 15 '22 at 19:48
Well, if $g$ is periodic of period $1$, then that will follow. – John Wayland Bales Aug 15 '22 at 21:12