From the limit of the derivative, infer the behavior of the antiderivative

[SchroedingersLion]

Greetings!

In statistical mechanics, when studying diffusion processes, one often finds the following reasoning:

Suppose there is a strictly positive differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ with $\lim_{x \rightarrow +\infty} {f'(x)} = a > 0$.
Then for sufficiently large $x$, we have $f(x) \sim ax$, that is $f(x)\in O(x)$, using the big-O notation.

I cannot find a rigorous proof for this. I want to show that there is $x_0 \in \mathbb{R}$ and $M>0$ such that for $x>x_0$ we have
$$
\frac{f(x)}{|x|} \le M.
$$

Any hints?

SL

 

[fresh_42]

This is one of these questions where the obvious truth makes it difficult to formalize it. I'm not sure, but I give it a try:
\begin{align*}
f(x_0 + v) - f(x_0) &= av + o(v)\\
f(x) &= f(x_0) + a (x-x_0) +o(x-x_0)= ax + C \\
\dfrac{f(x)}{x}&\leq a+\dfrac{C}{x}\leq a+\dfrac{C}{x_0}=:M
\end{align*}

 

[Office_Shredder]

Here's something way more rigorous. The key is to not try to get the slope to be a, it's too hard (and I'm not sure it's even true)

From the limit there exists some point w such that $f'(w) \leq a+1$ for all $x\geq w$. Then $f(x) \leq f(w)+ (a+1)(x-w)$ for all $x\geq w$
Obviously you can use a+anything, or even something like 1.01a if you're looking for a relative slope instead.

 

[fresh_42]

Here's something way more rigorous. The key is to not try to get the slope to be a, it's too hard (and I'm not sure it's even true)

From the limit there exists some point w such that $f'(w) \leq a+1$ for all $x\geq w$. Then $f(x) \leq f(w)+ (a+1)(x-w)$ for all $x\geq w$
Obviously you can use a+anything, or even something like 1.01a if you're looking for a relative slope instead.

But that's basically the same as I wrote, only with the safety margin $a+\varepsilon$ instead.

 

[Office_Shredder]

$\dfrac{f(x)}{x}\leq a+\dfrac{C}{x}$

I think this line is false, because you elided over what o(x) means. This directly transforms into $f(x)\leq ax+C$, but if I define $f(x)=ax+ \frac{x}{ln(x)}$ for $x\geq 2$, then I get the derivative limit we want but that inequality doesn't hold.

Also, I suspect the OP is looking for something that never writes down things like $o(x)$ and then throws it out, because throwing it out assumes results like what is asked in the first post.

 

[fresh_42]

This was the lazy version of $\lim_{x \to x_0}\dfrac{r(x-x_0)}{|x-x_0|}=0$. That all goes down in $C$, as it can be replaced by e.g. $1$ if we divide by the large $x$ anyway.

 

[Infrared]

The key is to not try to get the slope to be a, it's too hard (and I'm not sure it's even true).

Could you explain what you mean by this?

 

[fresh_42]

Could you explain what you mean by this?

I think that the statement only allows to conclude about the asymptotic behavior. E.g. $f$ could be a damped sine curve along $x\longmapsto ax$, which is only periodically an actual slope. But $a\pm \varepsilon$ for large $x$ should do the job.

 

[Office_Shredder]

The way I was thinking of it is that it's tempting to think f(x) can be bounded above by some line of slope a if you just shove the line high enough up on the graph, but that's not actually possibly. I think $ax + x/ \ln(x)$ is a great function to think about here, it's basically $(a+\epsilon)x$ for $\epsilon >0$ but eventually is strictly above any line with slope a.

 

[MathematicalPhysicist]

It follows from Lagrange intermediate theorem, take $x>0$, then there exists $0<c_x<x$, such that:
$(f(x)-f(0))/x = f'(c_x)$, now let $c_x \to \infty$ then $x\to \infty$ and we get: $(f(x)-f(0))/x\to a$, but then $f(0)/x \to 0$ as $x\to \infty$, so $f(x)/x \to a$.

 

[Office_Shredder]

Why are you allowed to let $c_x\to \infty$? You have no control over how that number is picked to begin with.

 

[MathematicalPhysicist]

Why are you allowed to let $c_x\to \infty$? You have no control over how that number is picked to begin with.

Back to the drawing board...

 

[WWGD]

I haven't thought about it too carefully, but why not use the naive approach:

If $lim _{x\ rightarrow \infty}f'(x)=a>0$

Then there is M with:

$f'(x)-a < \epsilon$ for x>M

Then we integrate both sides. Pretty sure we need more than the continuity of f ( which follows from the differentiability of f ) , e.g., absolute continuity ( a.c) of f to conclude $\int f'(x)=f(x)+C$; C a constant. In that case, we are done. Cantor Ternary function is the standard example ( iirc) of a function that is continuous but not a.c . I wonder if the conditions of the problem are enough for f to be a.c.

 

[fresh_42]

Integration yields only $\Delta f$. You need a small $\Delta x$, which is basically the same thing as I wrote in post #2. I only forgot to substitute $f'(x_0)$ by $a+\varepsilon$ instead of $a$.

 

[WWGD]

Integration yields only $\Delta f$. You need a small $\Delta x$, which is basically the same thing as I wrote in post #2. I only forgot to substitute $f'(x_0)$ by $a+\varepsilon$ instead of $a$.

I think if f is absolutely continuous, integrating f' ( indefinite) returns f, no? I think that is the defining property of a.c

 

[fresh_42]

I don't know what absolute continuous means. I thought it meant continuity of $|f|$. But $f$ is positive anyway.

All we have is the FTC, no indefinite integral. It is anyway merely another notation of the Weierstraß formulation. Write down the definition of differentiation: $f(x)=f(x_0)+(a+\varepsilon )(x-x_0)+r(x-x_0)$, with $x_0$ large enough such that $f'(x_0)=a+\varepsilon$, divide the entire thing by $x$ for $x>x_0$, and $\dfrac{f(x)}{x} < M$ pops up.

 

[WWGD]

I don't know what absolute continuous means. I thought it meant continuity of $|f|$. But $f$ is positive anyway.

All we have is the FTC, no indefinite integral. It is anyway merely another notation of the Weierstraß formulation. Write down the definition of differentiation: $f(x)=f(x_0)+(a+\varepsilon )(x-x_0)+r(x-x_0)$, with $x_0$ large enough such that $f'(x_0)=a+\varepsilon$, divide the entire thing by $x$ for $x>x_0$, and $\dfrac{f(x)}{x} < M$ pops up.

A.C is sort of continuity in terms of measure. For every $\epsilon>0$ there is$\delta>0$ so that. If $\Sigma_{k=1} ^n |x_k-x_j|<\delta$ then $\Sigma | f(x_k)-f(x_j)|< \epsilon$ for non-overlapping intervals $(x_k, x_j)$. But your method works so maybe better that than bringing more complicated machinery.

 

[SchroedingersLion]

I would not have thought that this is so non-trivial. Thanks for all the responses.
In case anyone is curious, here is a screen of the paper where I found this reasoning - eqs. (2.3)-(2.4):

Spoiler

Source: Wood & Lado: "Monte Carlo calculation of normal and abnormal diffusion in Ehrenfest's wind-tree model"
https://www.sciencedirect.com/science/article/pii/0021999171901094

 

[SchroedingersLion]

I don't know what absolute continuous means. I thought it meant continuity of $|f|$. But $f$ is positive anyway.

All we have is the FTC, no indefinite integral. It is anyway merely another notation of the Weierstraß formulation. Write down the definition of differentiation: $f(x)=f(x_0)+(a+\varepsilon )(x-x_0)+r(x-x_0)$, with $x_0$ large enough such that $f'(x_0)=a+\varepsilon$, divide the entire thing by $x$ for $x>x_0$, and $\dfrac{f(x)}{x} < M$ pops up.

@fresh_42 , since you answer seems to have been agreed upon, can you explain what happens with the term $r(x-x_0)$?
The Taylor series would go $f(x)=f(x_0)+(a+\varepsilon )(x-x_0)+\frac 1 2 f''(x_0)(x-x_0)^2 + ...$, right?
Now, dividing by x leads to terms proportional in x so that the right-hand side would explode upon taking the limit.

 

[fresh_42]

Weierstraß has the advantage over Taylor, that we do not have to think about how many terms there are, where it converges or not, and how the remainder term looks like. Differentiation alone says that
$$
f(x_{0}+v)=f(x_{0})+f'(x_0)(v)+r(v)
$$
where the remainder converges faster to zero than $v$ does: $\lim_{v \to 0}\dfrac{r(v)}{v}=0.$

Now all we have to do is replace $f'(x_0)$ by $a+\varepsilon$ since we do not know the exact slope, only that it is near $a$ for large $x_0$, then replace the direction of differentiation by $v=x-x_0$ for $x>x_0$, use the linearity of the derivative to write $f'(x_0)(x-x_0)=f'(x_0)x-f'(x_0)x_0,$ divide by $x$, and finally replace small terms by a constant like $1$, and sort all constant terms (those without an $x$) and call them $M$.

If you use Taylor, then you have to deal with the complete theorem and its conditions. However, this is not necessary as we only need the first derivative, i.e. the definition of differentiation.