## Analysis – Increasing functions

Some of the topics discussed today in lecture came up on a question at Math.Stackexchange. I am reposting my answer here, slightly edited.

The question, by Mah Moud, was whether there can be a function whose derivative at a point is positive but the function is not increasing.

This is actually a somewhat subtle issue.

Suppose first that $I$ is an interval on $\mathbb R$, that $f:I\to\mathbb R$, that $f'(a)>0$, and that $a$ is an interior point of $I$. We could just consider one-sided derivatives if $a$ is an end-point of $I$, but that seems an unnecessary distraction.

That $f'(a)>0$ means that $\displaystyle \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}>0,$

so if $h\ne0$ is sufficiently small, we can assume both that $a+h\in I$, and that $\displaystyle \frac{f(a+h)-f(a)}{h}>0.$

Considering $h>0$, this means that if $b>a$ is sufficiently close to $a$, then $f(b)>f(a)$. Considering $h<0$, this means that if $b is sufficiently close to $a$, then $f(b). If this is all we mean by “ $f$ is increasing at $a$” then indeed $f'(a)>0$ implies that. Similarly, $f'(a)<0$ would imply that $f$ is decreasing at $a$.

The definition here would be that $k$ is increasing at $a$ iff there is an interval $I$ about $a$ such that $k(b) if $b and $b\in I$, and $k(a) if $b>a$ and $b\in I$. But, really, this is a silly notion: The function below given by $\displaystyle k(x)=\left\{\begin{array}{cl}0&\mbox{ if }x=0,\\-x-3&\mbox{ if }-2\le x<0,\mbox{ and }\\-x+1&\mbox{ if }0

is increasing at $0$. However, what is not true is that if $f'(a)>0$, then $f$ is increasing on an interval around $a$ (even if we assume that $f$ is differentiable everywhere). It is common to define the notions of increasing and decreasing so that they apply to functions defined on intervals, rather than to individual points of the domain: We say that $f$ is increasing on a set $A$ iff whenever $x are in $A$ then $f(x)\le f(y)$ (with $\le$ replaced by $<$ if we insist that “increasing” be interpreted in the strict sense.)

To see that indeed $f$ needs not be increasing on any interval containing $a$, no matter how small, consider the example suggested in these slides by Louis A. Talman, on The Mother of All Calculus Quizzes: Let $g(x)$ be the function given by $g(0)=0$ and, if $x\ne0$, then $\displaystyle g(x)=\frac x2+x^2\sin\left(\frac1x\right).$

This function is differentiable everywhere, with $\displaystyle g'(0)=\frac 12+\lim_{h\to 0}\frac{h^2\sin(1/h)}h=\frac12$, and $\displaystyle g'(x)=\frac12+2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right)$

for $x\ne0$. Note that there are (both positive and negative) values of $x$ arbitrarily close to $0$ where $g'(x)<0$, so “ $g$ is decreasing at $x$“. This is because we can find arbitrarily small $x$ with $\cos(1/x)=-1$, so $\sin(1/x)=0$ and $g'(x)=-1/2<0$. (Note that the $x/2$ in the definition of $g$ could be replaced with any function with small enough derivative to ensure the same behavior.) Finally, note that $g'$ is not continuous at $0$. This is an essential feature of the example. For suppose that $f'$ exists on a neighborhood of $a$ and is continuous at $a$. If $f'(a)>0$, then for $b$ sufficiently close to $a$ we have $f'(b)>0$ as well. This means that $f$ is indeed increasing on a neighborhood of $a$. For example, if $x and they are close enough to $a$ to ensure that $f'(t)>0$ at all $t\in[x,y]$, then use the mean value theorem to see that there is a $z\in(x,y)$ with $\displaystyle f(y)-f(x)=f'(z)(y-x)>0,$

that is, $f(x).

### One Response to Analysis – Increasing functions

1. andrescaicedo says:

Related: Jack B. Brown, Udayan B. Darji, and Eric P. Larsen. Nowhere monotone functions and functions of nonmonotonic type, Proceedings of the American Mathematical Society, 127 (1), (1999), 173-182. MR1469402 (99b:26015).

From the abstract:

We investigate the relationships between the notions of a continuous function being monotone on no interval, monotone at no point, of monotonic type on no interval, and of monotonic type at no point. In particular, we characterize the set of all points at which a function that has one of the weaker properties fails to have one of the stronger properties.