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I have a function $f:\mathbb R\to \mathbb R$ and $f$ doesn’t have any restriction, so it might not be continuous and so on. If the condition $$f(x+1)-f(x)=c$$ holds $\forall x\in \mathbb R$, does that mean $f$ is linear? If not why this textbook say this? enter image description here

This is a solution to the IMO Shortlist 2015 A2 which says Find all $f:\mathbb Z \to \mathbb Z$ s.t. $$f(x-f(y))=f(f(x))-f(y)-1$$

PNT
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1 Answers1

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This answer will rather be a compilation of the comments. The short answer is no.

First, let's take a look at the case $c = 0$. Then, $f(x+1)-f(x) = 0,\ \forall x\in\mathbb R,$ which is a definition of a periodic function with period $1$. In the comments there are couple of suggestions:

  1. $f(x) = \sin (2\pi x)$ which is even smooth, but still not linear. (David Mitra)
  2. $f(x) = \begin{cases}1,\, x\in\mathbb Z\\ 2,\,\text{otherwise}\end{cases},$ which is not continuous, yet periodic, since $x+1$ and $x$ are either both integers or both non-integers and therefore $f(x+1) = f(x).$ (TheBestMagician)
  3. $f(x) = x - \lfloor x \rfloor$ which gives us the fractional part of real $x$. Its graph is sawtooth wave. It is periodic with period $1$ since $\lfloor x + 1 \rfloor = \lfloor x \rfloor +1$ for any real $x$.

For the case when $c\neq 0$, the easiest counterexample that came to mind and which I mentioned in the comments is $f(x) = c\lfloor x \rfloor$. Using the fact $\lfloor x + 1\rfloor = \lfloor x \rfloor + 1$ again, you can easily check that $f(x+1)-f(x) = c$. If $c = 1$, you can take a look at the graph of $f$ here.

To find more counterexamples, we can follow hint from Will Jagy in the comments and define function $g(x) = f(x)-cx$. It is easy to check that $g(x+1)-g(x) = 0$ and therefore $g$ is periodic function with period $1$. Therefore, all $f$ that satisfy your condition are of the form $$f(x) = cx + g(x)$$ where $g$ is periodic with period $1$. We can also conclude that $f$ is linear if and only if $g$ is constant. Since almost all periodic functions are not constant, almost all $f$ are nonlinear. This is also suggested in the comment by Thomas Andrews since for any $g$ defined on $[0,1)$ you can define $\tilde{g}(x) = g(x-\lfloor x\rfloor)$ and $\tilde g$ will be periodic with period $1$.

It's perhaps worth noting that $f(x+h)-f(x) = c$ will similarly fail for any fixed $h$, you just replace periodic functions with period $1$ with periodic functions with period $h$ in the above discussion. If you wanted to make sure $f$ is linear, you need much stronger condition like $$f(x+h)-f(x) = ch,\ \forall x\in\mathbb R,\, \forall h\in\mathbb R$$ which will make sure that $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = c,$$ i.e. $f$ is differentiable on all $\mathbb R$ and its derivative is constant. This does imply that $f(x) = cx + d$ for some constant $d$.

EDIT: Or, simply note that $f(x+h)-f(x) = ch$ for all $x,h\in\mathbb R$ implies $f(h)-f(0) = ch$ for all $h\in\mathbb R$.

Ennar
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  • Can you see the edit? – PNT Jan 05 '22 at 18:29
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    @Yassir, it's simple: functions $f\colon \mathbb Z \to \mathbb Z$ are not the same as $f\colon\mathbb R\to\mathbb R$. If $f(x+1)-f(x) = c$, you can prove by induction $f(x+n)-f(x) = nc$ for all $n\in\mathbb Z$. Now let $x = 0$. Note the similarity between $f(x+n)-f(x) = nc$ for all $n\in\mathbb Z$ and what I mention in my answer $f(x+h)-f(x) = ch$ for all $h\in\mathbb R$. – Ennar Jan 05 '22 at 18:40
  • @Yassir, just for the future, what you did is ask one thing, when in reality you wanted to know something else. This can be problematic since editing questions in such cases can easily invalidate answers, and therefore someone's effort. The best way to avoid this is to post the full problem right from the start and point to where your confusion lies. – Ennar Jan 05 '22 at 18:52