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 $f(x)=ax+b$?
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2Here 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 . – Alphanerd Jul 22 '18 at 11:22
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You've shown that $\lim f''(x) = 0$, not that $f''(x) = 0$. – Kenny Lau Jul 22 '18 at 11:23
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Oh I see , is there any way to conclude $f''(x)=0$ from the limit expression, using my approach ? . – Alphanerd Jul 22 '18 at 11:25
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Not sure if it helps, but we may write $\int e^x f(x+1)dx=\int e^x[f'(x)+f(x)]dx=e^x f(x)+M$ – Mythomorphic Jul 22 '18 at 11:50
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2Also related: Existance of a Twice Differentiable Function, and one which actually gives a sufficient condition so that we can conclude $f(x) = ax + b$ is Finding all differentiable functions satisfying a property. – mechanodroid Jul 22 '18 at 12:05
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2@mechanodroid TLDR, what's the solution to OP's question ? OP requires that $f'$ have a limit at $\infty$. – Gabriel Romon Jul 22 '18 at 12:10
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4I don't see an answer to this question in any of the posts quoted above. – Kavi Rama Murthy Jul 22 '18 at 12:11
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3Note, the TITLE of this is a duplicate, but the actual question in the text is not. – GEdgar Jul 22 '18 at 12:25
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5@GEdgar why did you vote to close then ? – Gabriel Romon Jul 22 '18 at 12:27
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The MVT tells us $f'(x)$ is constant on the interval $(x,x+1)$ for all $x \in \mathbb{R}$ no? Is this sufficient to go all the way and say $f$ is a linear function? – Theo Diamantakis Jul 22 '18 at 12:39
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I then voted to re-open. – GEdgar Jul 22 '18 at 12:46
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3@hctb What did you try to solve this? – Did Jul 22 '18 at 13:23
3 Answers
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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 ($1$).
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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 $n \to +\infty$?
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Maybe $(x_n)$ is bounded above but then il has a limit $l_1$ such that $f'(l_1)=f'(x_0)$ and the same construction you did provide a strictly increasing sequence $(l_{1,n})$ such that $f'(l_{1,n}=f'(x_0)$. If $(l_{1,n})$ is bounded above likewise it has a limite $l_2$ and the same construction provides a sequence $(l_{2,n})$ such that $f'(l_{2,n}=f'(x_0)$ and so on,. can this be used to construct an unbounded strictly increasing sequence $(y_n)$ such that $f'(y_n)=f'(x_0)$ for any $n$? – Fred Bernard Aug 22 '24 at 10:46
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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 $\Large f(x)\text{ is linear}$
Mostafa Ayaz
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