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My problem is pretty straight-forward to state, but for the sake of completeness, I'll give a short explanation of how it appeared.

I came across a solution to problem 1 of IMO 2019 on Youtube. The problem is:

Find all functions $f\colon\mathbb{Z}\rightarrow\mathbb{Z}$, such that $f(2a)+2f(b)=f(f(a+b))$ for all $a,b\in\mathbb{Z}$.

The solution goes as follows:

First put $a=0$. Then for all $b$, we have

$$f(0)+2f(b) = f(f(b)).$$

Second put $a=1$. Then for all $b$,

$$f(2)+2f(b) = f(f(b+1)).$$

Use the first equation with $b+1$ in the second equation, i.e. by the first equation, we have $f(0)+2f(b+1)=f(f(b+1))$, so $$f(2) +2f(b) = f(0)+2f(b+1),$$ or $$\frac{f(2)-f(0)}{2} = f(b+1)-f(b)$$ for all $b\in\mathbb{Z}$.

Since the increments are constant, it follows that $f$ is linear. You can plug a linear expression into the original functional equation to check what fits.

My question is this: If we had instead had a real function $f$, and came to the conclusion that $$f(x+1)-f(x)=k$$ for all $x$ and some constant $k$, are there any necessary or sufficient conditions that I can impose on $f$ to make $f$ linear? Without any further restrictions, all I can say is that for any real number $r\in [0,1)$, and for any integer $n\in\mathbb{Z}$, we have

$$f(n+r) = an+f(r) = a(n+r) + (f(r)-ar),$$ and $f(r)-ar$ might be different from $f(0)$.

Source: The video in question is this one.

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You can define any function you like on an interval of length $1$ and then extend it using your $k$ by using the functional relation: $$f(x+1)=f(x)+k.$$

Inside the given interval, you of course won't have any linearity whatsoever

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Take$$f(x)=\begin{cases}1&\text{ if }x\in\mathbb Z\\0&\text{ otherwise.}\end{cases}$$Then $(\forall x\in\mathbb R):f(x+1)-f(x)=0$. However, $f$ is not linear.

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    $\begingroup$ I guess you could still say: Wherever it is continuous, it is linear $\endgroup$ – dbx Aug 13 at 19:21
  • $\begingroup$ What about $f(x)= \sin 2\pi x$, when $k=0$ ? $\endgroup$ – Fermat Aug 13 at 19:29
  • $\begingroup$ Sorry, it looks like my question was not entirely clear. I am asking: Given that $f(x+1)-f(x)$ is constant for all $x$, are there any conditions I can add to $f$ that forces $f$ to be linear (either sufficient or necessary)? I'll edit my question to make it clearer. $\endgroup$ – Mankind Aug 13 at 19:33
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I'll give you a continuous Example: $$f(x)=|\sin (\pi x)|.$$

In general, you just need a periodic function, with period 1, and that will satisfy your equation with $k=0$.

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Any function

$$ f(x) = k x +\Phi(x) $$

with $\Phi(x)$ periodic with period $1$ is a solution. For instance

$$ f(x) = k x + \sin(2\pi x) $$

is such a function and $f(x)$ is not linear.

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If we consider $x=\left \lfloor x \right \rfloor + \left \{ x \right \}$ and $x>0$ then $$f(x)-f(\left \{ x \right \})= f(\left \lfloor x \right \rfloor + \left \{ x \right \})-f(\left \{ x \right \})=\\ f(\left \lfloor x \right \rfloor + \left \{ x \right \})- f(\left \lfloor x \right \rfloor -1 + \left \{ x \right \})+ f(\left \lfloor x \right \rfloor-1+\left\{ x \right \})- f(\left \{ x \right \})=\\ k+f(\left \lfloor x \right \rfloor-1+\left\{ x \right \})- f(\left \{ x \right \})=\\ 2k+f(\left \lfloor x \right \rfloor-2+\left\{ x \right \})- f(\left \{ x \right \})=...= \left \lfloor x \right \rfloor k$$ or $$f(x)=\left \lfloor x \right \rfloor k + f(\left \{ x \right \}) \tag{1}$$ At this point it becomes obvious that

$f(x)$ is linear in $\mathbb{R} \iff f(x)$ is linear in $[0,1)$ with $k$ as the slope.

$\Leftarrow$ Obviously, if $f(x)$ is linear in $[0,1)$ with $k$ as the slope, i.e. $f(x)=kx+b, x\in[0,1)$, then $$f(x)=\left \lfloor x \right \rfloor k + f(\left \{ x \right \}) = \left \lfloor x \right \rfloor k + \left \{ x \right \}k+b= (\left \lfloor x \right \rfloor+\left \{ x \right \})k+b=\\ xk+b, x\in\mathbb{R}$$ $\Rightarrow$ Now, if $f(x)$ is linear in $\mathbb{R}$ or $f(x)=ax+b$, then from $(1)$ $$f(1)=k+f(0) \tag{2}$$ and $$f(0)=a\cdot 0+b=b$$ and $f(1)=a+b$ and $f(1)=k+b$ or $k=a$ and result follows.

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