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From what I understand we say that a function is increasing on an interval $I$ if $$ x_1 < x_2 \quad\Rightarrow\quad f(x_1) < f(x_2). $$ for all $x_1,x_2\in I$. I understand that some might call this strictly increasing and allow for equality when saying increasing.

We know that when a function is differentiable on the interval $I$, then if $f'(x) > 0$ on $I$ then $f$ is increasing on $I$. But here we have to be careful because if $f(x) = x^3$ then $f$ indeed is increasing on $\mathbb{R}$ even though $f'(0) = 0$. So we can't conclude that if $f$ is increasing, then $f'(x)$ must be positive on the interval.

Also, even though $f(x) = x^2$, then $f$ is increasing on $[0,\infty)$ even though $f'(0) = 0$.

My question is

Does it make sense to talk about a function being increasing at a point?

The only definition I can think of making is to say that $f$ is increasing at a point $a$ if $f'(a) > 0$. But the problems with saying this is that then the function given by $f(x) = x^3$ is not increasing at $0$ even thought it is increasing on all of $\mathbb{R}$.

I am aware (from here) of the example with $$ f(x) = \begin{cases}x^2\sin\left(\frac{1}{x}\right) + \frac{x}{2} & \text{if }x\neq 0 \\ 0 & \text{if } x = 0\end{cases}. $$ This function is differentiable at $0$ with $f'(0) = \frac{1}{2} > 0$. For $x\neq 0$ we have $f'(x) = 2x\sin(1/x) - \cos(1/x) + 1/2$. And so for all $x_k = \frac{1}{2k\pi}$ we have $f'(\frac{1}{2k\pi}) = -\frac{1}{2} < 0$. That is, there is not interval around $0$ where $f$ is increasing. And so it shouldn't really make sense to say that $f$ is increasing at $0$.

One example I can think of where it maybe should make sense to take about a function being increasing at a point is when when we take the derivative of function $s(t)$ that gives the position of a particle. The derivative is then the velocity $v(t) = s'(t)$. And maybe this is just a language issue, but here it sounds reasonable saying that if $v(a)>0$ then the velocity is positive at time $t=a$ and so the rate of change is positive at $a$. This is almost saying that the position is increasing at that point.

Again, my question is just if it makes sense to talk about a function being increasing (or decreasing) at a point. If so, how is this defined?

user182601
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Thomas
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    I guess it makes sense to say around a point. – AnilB Apr 17 '13 at 16:08
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    @AnilBaseski: Yes, I agree that it makes sense to say that a function is increasing around a point if by that we mean that there is some interval around the point where $f$ is increasing. – Thomas Apr 17 '13 at 16:09
  • The derivative can be defined as the rate of change of f relative to a change of x. If we say that x is fixed then the definition is non-sense. I guess in the definition it is assumed that x is not fixed, which follows x is not a point. – AnilB Apr 17 '13 at 16:12
  • First off be careful with your example $f(x)=x^3$. You are using the first derivative test to characterize the notion of increase at a critical point. That's not allowed. The derivative is just a tool to help investigate the nature of a function, it can't be used to define the notion of increasing. Considerthe function where $f(x) = x+1$ if $x \in \mathbb{Q}$ and $f(x) = 0$ when $x \in \mathbb{R}/\mathbb{Q}$. This is increasing and not differentiable. – Wintermute Apr 17 '13 at 16:25
  • @mtiano: But that is my question. In my question I note that it doesn't really make sense to say that a function is increasing at a point if the derivative at the point is positive. So my questions is: Can we make sense of it? Also, in your example I guess that the function isn't increasing? ($f(1) = 2 > f(\pi) = 0$) – Thomas Apr 17 '13 at 16:27
  • Did you realize that your given piece wise function is indeed differentiable at x=0, but the derivative is not continuous at x=0? – imranfat Apr 17 '13 at 16:36
  • @imranfat: Yes, I did notice that. I guess that is the point about not defining a function to be increasing at a point when the derivative is positive. The example shows that in this case there is not small interval around $0$ where the function is increasing (the derivative is not continuous). – Thomas Apr 17 '13 at 16:40
  • Yep, that continuity of the derivative should not be taken lightly :))) – imranfat Apr 17 '13 at 16:45
  • @imranfat: True. It is too often just assumed :) – Thomas Apr 17 '13 at 16:45
  • Take a look at this. I do not think there is any obvious way to define increasing at a point other than what you have already discussed in your question. – Caleb Stanford Apr 17 '13 at 16:55
  • Doesn't the PDF (Talman, On Functions “Increasing at a Point”) you linked to already suggest (p. 3, first paragraph) the definition that Did gave below? (And it also gives a really good discussion of the issues involved.) – user182601 Feb 24 '24 at 07:51

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A function $f$ is increasing at $x$ if $f(t)\gt f(x)$ for every $t\gt x$ close enough to $x$ and $f(t)\lt f(x)$ for every $t\lt x$ close enough to $x$. More rigorously, one asks that there exists $\varepsilon\gt0$ such that, for every $(t,s)$ such that $x-\varepsilon\lt t\lt x\lt s\lt x+\varepsilon$, $f(t)\lt f(x)\lt f(s)$.

No notion of differentiability is needed. Consider for example the function $f$ defined by $f(t)=2t$ for every rational $t$ and $f(t)=t$ for every irrational $t$. Then $f$ is increasing at $x=0$ and only at $x=0$ while $f$ is nowhere differentiable.

Did
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  • Great. That definition makes sense. Do you know of any text book that uses this definition? – Thomas Apr 17 '13 at 16:46
  • This definition, surely with @Douglas’s emendation, probably is the only one that makes sense, but I don’t believe I’ve ever seen it before. It prompts the immediate question of whether a function can be increasing at every point of an interval without being increasing on that interval. – Lubin Apr 17 '13 at 16:50
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    Any textbook? I am afraid not. But the definition is natural to anybody studying the pathwise properties of random processes, most notably standard Brownian motion. – Did Apr 17 '13 at 17:00
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    @Lubin (and others): See my 23 April 2006 AP-Calculus post archived at Math Forum for a reasonably complete discussion of these matters, as well as my 16 April 2010 AP-calculus post also archived at Math Forum. – Dave L. Renfro Apr 17 '13 at 18:12
  • I can think of at least one more definition that makes sense, namely that $f$ is increasing on every sufficiently small interval around $x$. But as I haven't seen either definition in literature, I guess I'll defer to the one that makes sense in the Brownian motion context. – MartianInvader May 25 '16 at 22:33
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    Example Textbook: Sohrab's Basic Real Analysis. It has following definition in Problem 3.6: A function $f\in \mathbb{R}^{\mathbb{R}}$ is increasing at a point $x_0$ if there is a $\delta > 0$ such that $f(x)\leqslant f(x_0)\leqslant f(y)$ for all $x\in(x_0-\delta,x_0)$ and $y\in (x_0,x_0+\delta)$. – PeptideChain Sep 11 '16 at 13:20
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    This definition is also given in Michael Spivak's Calculus (fourth edition) on page 217. – Joe Aug 21 '21 at 19:53
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    The Math Forum links I gave in a comment above no longer work, and their archive has been taken off the internet from what I understand. However, I think the 2006 post is what I've copied into this 16 May 2006 sci.math post archived by google (other posts in that thread may also be of interest). See also this 29 December 2008 sci.math post. – Dave L. Renfro Oct 30 '21 at 20:27
  • @Thomas: Landau, Differential and Integral Calculus (1934, 1950 translation), p. 88, Definition 27. // Similar: Hardy, A Course of Pure Mathematics (1967 10e), p. 178, "Steadily increasing or decreasing functions" – user182601 Feb 24 '24 at 07:34