# $f(x+1)-f(x)=f'(x)$: prove $f(x)$ linear function [closed]

If I have a differentiable function $f:\mathbb{R}\to\mathbb{R}$ satisfies $f(x+1)-f(x)=f'(x)$ and $\lim_{x\to\infty}f'(x)=A$. Can I show $f(x)=ax+b$?

• Here is my guess : You can differentiate the given expression $f(x+1)−f(x)=f′(x)$ and then apply limit to the new expression(x tending to infinity) : $f'(x+1)−f'(x)=f′'(x)$ . You will get 0, from which we can conclude that f(x) has to be linear since its 2nd derivate is 0 . Jul 22 '18 at 11:22
• I don't see an answer to this question in any of the posts quoted above. Jul 22 '18 at 12:11
• Note, the TITLE of this is a duplicate, but the actual question in the text is not. Jul 22 '18 at 12:25
• @GEdgar why did you vote to close then ? Jul 22 '18 at 12:27
• @hctb What did you try to solve this?
– Did
Jul 22 '18 at 13:23

Suppose $f$ is differentiable, $$f(x+1)-f(x)=f'(x) \tag{1}$$ for all $x$, and $\lim_{x \to +\infty} f'(x) = A$.

I claim that $f'$ is constant, and therefore that $f$ has the form $ax+b$.

Suppose, for purposes of contradiction, that $f'$ is not constant. Then there is $x_0$ such that $f'(x_0) \ne A$. Take the case $f'(x_0) > A$. [The other case $f'(x_0)<A$ is done the same way.]

Differentiate the equation $f(x+1)-f(x)=f'(x)$ to conclude that $f''$ exists and that $f'$ is continuous. Function $f'$ achieves a maximum value $B > A$ on $[x_0,+\infty)$. The set where $f'(x)=B$ is nonempty, closed, and bounded above. Let $x_1 \in [x_0,+\infty)$ be such that $f'(x_1) = B$ and $f'(x) < B$ for all $x \in (x_1,+\infty)$.

Now note $f'(x) < B$ on $(x_1,x_1+1)$, so $f(x_1+1) - f(x_1) = \int_{x_1}^{x_1+1} f'(x)\;dx < B = f'(x_1)$. This contradicts ($1$).

Define $$g(x)=f(x)+ax+b$$therefore by substitution we get $$g(x+1)=g(x)$$which means that $g(x)$ is periodic with period $1$. Also $g'(x)=f'(x)+a$ and therefore has a limit in $\infty$. Since $g'(x)$ is also periodic this is possible only if it is constant over a period or over $\Bbb R$ because $$\lim_{x\to\infty}g'(x)=g'(0)\\\lim_{x\to\infty}g'(x+a)=g'(a)\\g'(a)=g'(0)$$for any $a\in [0,1]$. So we have $$g'(x)=c$$concluding that $$g(x)=cx+d$$which means that $$\Large f(x)=(c-a)x+d-b$$ or $\Large f(x)\text{ is linear}$

# A thought

Let $$x_0$$ be any real number. By the condtion, we have $$f(x_0+1)-f(x_0)=f'(x_0).\tag1$$ But by Lagrange mean value theorem, we may also obtain $$f(x_0+1)-f(x_0)=f'(x_1)(x_0+1-x_0)=f'(x_1),$$where $$x_0 Combining $$(1)$$ and $$(2)$$, we may claim there exists $$x_1 \in (x_0,x_0+1)$$ such that $$f'(x_0)=f'(x_1).$$

Consider $$x_1$$. By the condition, we also have $$f(x_1+1)-f(x_1)=f'(x_1).\tag3$$ Likewise, we may also claim there exists $$x_2 \in (x_1,x_1+1)$$ such that $$f'(x_1)=f'(x_2).\tag4$$

Now, repeat the process above. You may find a sequence of $$x_0 such that $$f'(x_0)=f'(x_1)=f'(x_2)=\cdots=f'(x_n).$$ Obviously, if we may prove that $$x_n \to +\infty$$ as $$n \to +\infty$$, then by the fact that $$\lim\limits_{x\to\infty}f'(x)=A,$$ we have $$\lim\limits_{n\to+\infty}f'(x_n)=A.$$ But $$f'(x_n)$$ is a constant sequence, thus $$f'(x_0)=f'(x_1)=f'(x_2)=\cdots=f'(x_n) =A.$$

Recall the arbitrariness of $$x_0$$. We may claim that $$f'(x)\equiv A,~~~~\forall x \in \mathbb{R},$$ which implies that $$f(x)=Ax+b$$ for all $$x \in \mathbb{R}.$$

But can we prove $$x_n \to +\infty$$ as $$n \to +\infty$$?