$f:\mathbb R \rightarrow \mathbb R$ be differentiable function such that $f'(x)$ is continuous and $f(x+1)=f(x)+1$ for all $x \in\mathbb R$.

Question

I was thinking about the problem:

Let $f:\mathbb R \rightarrow \mathbb R$ be differentiable function such that $f'(x)$ is continuous and $f(x+1)=f(x)+1$ for all $x \in\mathbb R$. Then which of the following options is correct?

(a) $f'(x)$ must be bounded,

(b) $f(x)$ must be bounded,

(c) Both $f(x)$ and $f'(x)$ must be bounded,

(d) Both $f(x)$ and $f'(x)$ must be unbounded.

My attempts: If I take $f(x)$ to be of the simplest form that is $f(x)=x,$ so that the given condition is satisfied then we see that $f'(x)=1,$ which is bounded. So I think that choice (a) is the correct option. Am I going in the right direction? I want a proof in a more generalized way. Please help. Thanks in advance for your time.

Answer 1

Hint: Consider the properties of $g(x)=f(x)-x$.

Answer 2

By induction, it can be shown that $f(x+N) = f(x)+N \quad \forall N \in \mathbb{N}$.

So, $f$ has to be unbounded on $\mathbb R$.

On the other hand, $f'(x+1)= f'(x) \quad \forall x \in \mathbb{R}$. So, $f'(x)$ is a periodic function with period 1. If we know the behavior of $f'(x)$ for any interval of unit length, say $[0,1)$, we know its behavior on entire $\mathbb R$.

As $f'(x)$ is continuous, we know that it will have a maximum and a minimum on every interval of finite length.

$$\max_{x \in \mathbb R} f'(x) = \max_{x \in [0,1)} f'(x) \leq f'(\xi_1) \mbox{ for some } \xi_1 \in [0,1) $$

$$\min_{x \in \mathbb R} f'(x) = \min_{x \in [0,1)} f'(x) \geq f'(\xi_2) \mbox{ for some } \xi_2 \in [0,1) $$
So, $f'(x)$ is bounded.

Answer 3

$f(x)=x$ is a possibility. So you may remove $(b), (c), (d)$ at once.