Let $f:\mathbb{R}_+\rightarrow\mathbb{R}_+$ be a monotonic nonnegative non-decreasing $C^1$ divergent function, with $f'$ being monotonic nonnegative and bounded. I'm trying to prove that it is not possible to exist strictly increasing divergent real sequences $(a_n)_n$ and $(b_n)_n$ such that

$$ \lim_{n\to\infty} \frac{f(a_n)}{a_n^{1/3}} = \lim_{n\to\infty} \frac{f(b_n)}{b_n^{2/3}} = 1. $$

The only thing I've thought was to take a strictly increasing divergent sequence $(c_n)_n$ with $c_k \in (a_n)_n$ when $k$ even and $c_k \in (b_n)_n$ when $k$ odd, and to use the Mean value Theorem to try to control the growth of $f'$, but it was not very effective. I would appreciate any suggestion.


2 Answers 2


Lemma: Given $a<b$ and $A<B$ there exists a $C^\infty$ function $f$ such that all derivatives of $f$ vanish at $a,b,f$ is strictly increasing on $[a,b]$ and $f(a)=A, f(b)=B.$

Using the lemma, we can ping pong back and forth between the curves $x^{1/3}, x^{2/3}.$ We first move up from $x^{1/3}$ to $x^{2/3}.$ Now we hold $f$ constant until we run into the curve $x^{1/3}$ again. Then we can move back up to $x^{2/3}$ etc. Our function will be strictly increasing on the odd intervals obtained this way, and constant on the even intervals. The vanishing of all derivatives at each of $a_n,b_n$ insures the completed function will be $C^\infty.$

Added: I've been reminded by two commenters that $f'$ was itself supposed to be monotonic. I'll leave this post anyway.

Second answer: I believe there is such an $f.$

If we take a point $(x_0,y_0)$ on $y=x^{2/3},$ then the horizontal line $y=y_0$ will intersect $y=x^{1/3}$ at some point to the right. It follows that for sufficiently small $m>0,$ the line $y=y_0+m(x-x_0)$ will also intersect $y=x^{1/3}.$ Furthermore the continuation of any such line will again intersect $y=x^{2/3}.$ By taking $m$ small enough, we can assure this second point on $y=x^{2/3}$ is as far to the right as we want. Having obtained this second point, we can repeat the above. But this time we need to be sure the slope of the line is no larger than the first slope chosen. Since we have the lattitude of all sufficiently small slopes, this is no problem.

In this way we choose a sequence of points $b_n \to \infty$ such that the chords from $(b_n,b_n^{2/3})$ to $(b_{n+1},b_{n+1}^{2/3})$ have decreasing slopes, and such that each of these chords intersects $y=x^{1/3}.$ The chord intersections generate the sequence $a_n.$

So now with these chords inscribed on the graph of $y=x^{2/3},$ we choose a $C^\infty$ function $f$ whose graph coincides with the above chords most of the time, but near the vertices, smoothly transitions from one chord to the next, all the while preserving the strictly decreasing monotonity of the slopes, i.e., $f''\le 0$ everywhere.

  • $\begingroup$ There's also a condition that $f'$ is monotonic non decreasing, which I don't think is satisfied in this example. $\endgroup$
    – Kitegi
    Commented May 16, 2015 at 23:16
  • 1
    $\begingroup$ Indeed, it looks like on the even intervals $f'$ is $0$, and on the odd ones it'll be positive... $\endgroup$
    – shalop
    Commented May 16, 2015 at 23:19
  • $\begingroup$ Thanks for the correction to the both of you. $\endgroup$
    – zhw.
    Commented May 16, 2015 at 23:25
  • 2
    $\begingroup$ I added a second answer that shows, hopefully, that there exists such an $f.$ $\endgroup$
    – zhw.
    Commented May 17, 2015 at 0:39
  • 1
    $\begingroup$ Ha, nice. So just choose a decreasing sequence of slopes in such a way as to hit the more slowly-growing curve at each stage.Then smoothly interpolate... Good stuff! +1. BTW, you wrote $f'<0$ everywhere but it should be that $f'$ is decreasing everywhere. $\endgroup$
    – shalop
    Commented May 17, 2015 at 0:47

Case 1: $f'$ non-decreasing, that means $f$ is convex.

Since, $f,f',f''$ are all non-negative, we have $(f^3)'=(3f^2\cdot f')$ which is non decreasing. That means $f^3$ is also convex.

So $x\to\displaystyle\frac{f^3(x)-f^3(0)}{x}$ is non-decreasing. So as $x\to\infty$, it must either go to $\infty$ or to a finite limit $l$. Since we know that $\lim_{n\to\infty}\displaystyle\frac{f^3(a_n)}{a_n}=1$, then $x\to\displaystyle\frac{f^3(x)}{x}$ goes to $1$ as $x\to\infty$

This gives us that $\lim_{n\to\infty}\displaystyle\frac{f^3(b_n)}{b_n^2}=0=1$

So it's impossible.

Case 2: $f'$ is non-increasing.

I'll show a counterexample if $f$ is not $\mathcal C^1$ everywhere. It's not necessary but it makes the proof simpler and it's possible (while troublesome) to come up with a $\mathcal C^\infty$ counterexample.

We'll now construct a counterexample to the original proposal, by defining the function $f$ and the sequences $a_n$ and $b_n$ (and another one, $k_n$) and maintaining $f(a_n)=a_n^{1/3}$ and $f(b_n)=b_n^{2/3}$

We'll choose $a_n=(b_n^{2/3}+1)^3$

If $0\leq x\leq 1,\ f(x)=x$

$f$ crosses $x^{2/3}$ at $b_1=1$

Suppose that we've defined $f$ on $[0,b_n]$. We'll now define it for $[0,b_{n+1}]$

We'll define $f$ on $[b_n,a_n]$ as the chord between $(b_n,b_n^{2/3})$ and $(a_n,a_n^{1/3})$

The slope of $f$ is positive ($f$ must grow to reach $a_n^{1/3}$) and less than it was for $x<b_n$ (otherwise, $f$ will "pierce" $x^{2/3}$ and will always be strictly greater than it and $x^{1/3}$)

We'll define $b_{n+1}$ as the x-coordinate of the point where the extension of the previous chord (to the right of $a_n$) intersects $x^{2/3}$. Such a point must exist because $ax+b$ is eventually greater than $x^{2/3}$.

We define $f$ like this by induction. Notice that $f(b_n)=b_n^{2/3}$ and $f(a_n)=a_n^{1/3}$

This is how the function will look like: plot1

A little zoomed out: plot2

  • 1
    $\begingroup$ You're dealing with the function $f^3(x)$, but I think the more relevant function would be $f(x^3)$, no? $\endgroup$
    – shalop
    Commented May 16, 2015 at 23:21
  • $\begingroup$ But $f'$ is not necessarily non-decreasing, it's just nonnegative, like $f(x) = \log{x}$ and $f'(x) = 1/x$. $\endgroup$
    – Alufat
    Commented May 16, 2015 at 23:23
  • $\begingroup$ @Alufat Didn't you say that $f'$ is monotonic? $\endgroup$
    – Kitegi
    Commented May 16, 2015 at 23:24
  • $\begingroup$ Yeap, but it can be monotonic decreasing :P $\endgroup$
    – Alufat
    Commented May 16, 2015 at 23:26
  • 1
    $\begingroup$ @Alufat Alright, I believe I found a counterexample. $\endgroup$
    – Kitegi
    Commented May 17, 2015 at 0:50

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