8
$\begingroup$

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 $f(x)=ax+b$?

$\endgroup$
12
  • 2
    $\begingroup$ 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 . $\endgroup$
    – Alphanerd
    Jul 22 '18 at 11:22
  • 4
    $\begingroup$ I don't see an answer to this question in any of the posts quoted above. $\endgroup$ Jul 22 '18 at 12:11
  • 3
    $\begingroup$ Note, the TITLE of this is a duplicate, but the actual question in the text is not. $\endgroup$
    – GEdgar
    Jul 22 '18 at 12:25
  • 5
    $\begingroup$ @GEdgar why did you vote to close then ? $\endgroup$ Jul 22 '18 at 12:27
  • 3
    $\begingroup$ @hctb What did you try to solve this? $\endgroup$
    – Did
    Jul 22 '18 at 13:23
15
$\begingroup$

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 ($1$).

$\endgroup$
1
$\begingroup$

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 $\Large f(x)\text{ is linear}$

$\endgroup$
0
$\begingroup$

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 $x_0<x_1<x_0+1.\tag 2$ 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 $$x_0<x_1<x_2<\cdots<x_n$$ 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 $n \to +\infty$?

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.