# When does the condition $f(x+1)-f(x)=k$ for all $x$ imply that $f$ is linear?

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.

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

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.

• I guess you could still say: Wherever it is continuous, it is linear – dbx Aug 13 at 19:21
• What about $f(x)= \sin 2\pi x$, when $k=0$ ? – Fermat Aug 13 at 19:29
• 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. – Mankind Aug 13 at 19:33

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 $$k=0$$.

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.

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.