Limit of the Derivative of an Increasing, Bounded-Above Function

Question

Let $f:(0,\infty) \to \mathbb{R}$ be a differentiable function, which is increasing and bounded above. Then does $\lim_{x \to \infty} f'(x)=0$?

If we assume that $\lim_{x \to \infty} f'(x)$ exists, then this is true by an argument using the mean value theorem: By assumption $L=\lim_{x \to \infty} f(x)$ exists and is finite, and then $0=L-L=\lim_{n \to \infty} f(n+1)-f(n)=\lim_{n \to \infty} f'(x_n)$ for some $x_n \in (n,n+1)$ by the mean value theorem. But this doesn't work if we don't assume $\lim_{x \to \infty} f'(x)$ exists because $x_n$ isn't an arbitrary sequence with $x_n \to \infty$.

Intuitively it seems that it should be true without this assumption, but of course that doesn't mean that it's true.

It is not true. Take any smooth approximation of $\arctan(\lfloor x\rfloor)$ in such a way that the derivative is unbounded over $\mathbb{N}$. — Jack D'Aurizio, Jan 01 '15 at 17:21
If $f(x)$ is increasing on $(0,\infty)$ then isn't $f'(x)>0$ for all $x$ in the bound? — Ari, Jan 01 '15 at 17:23

score 3 · Accepted Answer · 2015-01-01T18:02:49.483

3

That's not true. Let $g: \mathbb R\to (0,\infty)$ be a continuous function so that

$$\int_0 ^\infty g(x) dx < \infty,\ \ g(n) = 1 \text{ for all }n\in \mathbb N.$$

Define

$$f(x) = \int_0^x g(s) ds$$

then $f$ is increasing and bounded above but $f'(s)$ does not have a limit as $s\to +\infty$. (If limit exist, it has to be 1. But then $f$ would not be bounded above)

Remark to construct such a $g$, it suffices to construct $g$ on $[n, n+1]$ so that $g(n) = g(n+1) =1$ and $\int_n^{n+1} g(x)dx < \frac{1}{2^n}$. Let $h$ be defined by

$$h(x) = -2^{n+1} (x-n) + 1 \text{ on } [n, n+ 2^{-n-1}],$$ $$h(x) = 0 \text{ on } [ n+ 2^{-n-1}, n+1 - 2^{-n-1}],$$ $$h(x) = 2^{n+1} (x - n-1 + 2^{-n-1}) \text{ on }[n+1 - 2^{-n-1}, n+1]$$

then $h: [n, n+1] \to [0,1]$ is continuous and $\int_n^{n+1} h(x) dx = 2^{-n-1}$. Now let $0<\epsilon_n<1$ be so small such that

$$g(x) = \max\{ h(x) , \epsilon_n\}$$

satisfies $\int_n^{n+1} g(x) dx < 2^{-n}$.

edited Jan 01 '15 at 18:02

answered Jan 01 '15 at 17:28

1

How do you know that $g$ exists? – Zubin Mukerjee Jan 01 '15 at 17:37
I think I can picture such a $g$, but can one write down a formula for it? – user4601931 Jan 01 '15 at 17:37
@dmdmdmdmdmd It has to be very close to $0$ near every integer, and then spike up to $1$ at the integer value (with increasing narrowness as $n$ gets larger and larger) ... I'm not even sure if this can be continuous on all of $\mathbb{R}$ – Zubin Mukerjee Jan 01 '15 at 17:38
That's what I had in mind, and now I have the same question as you. – user4601931 Jan 01 '15 at 17:39
@dmdmdmdmdmd: Such a function exists: see the edit. – Jan 01 '15 at 18:02
@ZubinMukerjee: The answer is edited. Please have a look. – Jan 01 '15 at 18:03

Limit of the Derivative of an Increasing, Bounded-Above Function

1 Answers1