# Nowhere monotonic continuous function

Does there exist a nowhere monotonic continuous function from some open subset of $\mathbb{R}$ to $\mathbb{R}$? Some nowhere differentiable function sort of object?

The Weierstrass function, mentioned in other answers, is indeed an example of a nowhere monotone function, meaning that $f$, even though continuous and bounded, is increasing at no point, decreasing at no point (and differentiable at no point as well). Details of this can be found in Example 7.16 in van Rooij, and Schikhof, A second course on real functions, Cambridge University Press, 1982.

That $f$ is increasing at $a$ means that there is a neighborhood $I$ of $a$ such that if $t<a$ is in $I$, then $f(t)\le f(a)$, and if $t>a$ is in $I$, then $f(t)\ge f(a)$. Thus, $f$ not increasing at $a$ iff any neighborhood of $a$ has points $t$ such that $(f(t)-f(a))(t-a)<0$. Being decreasing at $a$ can be stated similarly. See here.

We know that if $f$ is differentiable at $a$ and $f'(a)>0$ then $f$ is increasing at $a$, and if $f'(a)<0$, then $f$ is decreasing at $a$, so if a nowhere monotone function has a point $a$ in its domain where $f'(a)$ exists, then we must have $f'(a)=0$. It is indeed possible for a non-constant continuous increasing function $f$ to satisfy $f'(a)=0$ almost everywhere (we say that $f$ is singular). (Of course, if $f'(a)=0$ everywhere, then $f$ is constant.) The best known example of this phenomenon is Cantor's function, also known as the Devil's staircase (The link goes to O. Dovgoshey, O. Martio, V. Ryazanov, M. Vuorinen. The Cantor function, Expositiones Mathematicae, 24 (1), (2006), 1-37).

The above being said, note anyway that being increasing at a point is far from being increasing in a neighborhood of the point. If we require that $f$ is differentiable (and not constant), then there will be points $a$ where $f'(a)>0$ (so $f$ is increasing at $a$) or $f'(a)<0$ (so $f$ is decreasing at $a$). Nevertheless, as shown for example in Katznelson and Stromberg (Everywhere differentiable, nowhere monotone, functions, The American Mathematical Monthly, 81, (1974), 349-353) we can still find differentiable functions $f$ that are monotone on no interval. (I briefly state some properties of their example here; there used to be an accessible link to the paper, but apparently that is no longer the case.)

• If a function is not dofferentiable, how can we define increasing/decreasing at a point? I understand monotone on an interval but not a point – Ovi Jul 20 '18 at 16:42
• The definition of being increasing at a point is given in my answer (second paragraph). I agree as a pointwise notion it is not very satisfying, see the link at the end of that paragraph. – Andrés E. Caicedo Jul 20 '18 at 16:46
• Ah so I think that's equivalent to saying $f$ is not monotone on any interval, right? – Ovi Jul 20 '18 at 16:52
• Yes, nowhere monotonicity is the same as not being increasing or decreasing at any point. – Andrés E. Caicedo Jul 20 '18 at 16:55

Check my answer to this question:

Is this condition sufficient to ensure monotonicity of a function?

For that function, there are enough details so you can prove the following:

For any $x \in (0,1)$ and any $\epsilon >0$ there exists a $y$ so that $0< |x-y| < \epsilon$ and $f(x)=f(y)$.

That function is continuous, nowhere differentiable and the above result shows that it cannot be monotonic at any point, since it is not locally constant.

The Weierstrass function is non-monotonic over any interval. I'm not sure you can prove it non-monotonic at every point.

• What do you mean by "at every point"? Monotonicity is, as far as I understand, a property over some open interval. If you mean "Has an environment such that..." or "Does not have an environment where it is monotonous" then an environment must contain an interval, since it is non-monotonic over every interval it has to be that no neighbourhood of any point is monotonic. – Asaf Karagila May 31 '11 at 14:03
• @Asaf: I was thinking one could say $f$ is monotonic increasing at $x_0$ if you can find $\epsilon$ such that $f(x) \lt f(x_0)$ for all $x \in (x_0-\epsilon,x_0)$ and $f(x) \gt f(x_0)$ for all $x \in (x_0,x_0+\epsilon)$. Maybe this isn't standard. – Ross Millikan May 31 '11 at 14:39
• So it is somewhat like a "left-minima" point... About standard terminology I can't say as I'm far far away from this field :-) – Asaf Karagila May 31 '11 at 15:38
• @Asaf: I've seen the terminology "point of increase" and "point of decrease". And, FWIW, a Brownian motion sample path gives an example with none of either. – Nate Eldredge May 31 '11 at 17:12
• I didn't realize this was from 2011 until I had looked up what I was going to mention. Although the participants probably know plenty about pointwise monotonicity at this point (pun intended), it would probably be useful for me to archive it here anyway. See Brown/Darji/Larsen's Nowhere monotone functions and functions of nonmonotonic type, Proceedings of the American Mathematical Society 127 #1 (January 1999), 173-182 for one literature entry point to these notions. – Dave L. Renfro Apr 15 '13 at 17:36

Every monotonic function is almost everywhere differentiable (Theorem 4.3 - it's due to Lebesgue), so as an example of nowhere monotonic function you can just take any nowhere differentiable function (for example mentioned above the Weierstrass function).

• Oh, I haven't noticed the question was asked almost two years ago, the last edit misled me. – Damian Sobota Apr 15 '13 at 17:31

I was writing up a "continuous" construction for another question that got deleted. Though very belated, this idea seemed sufficiently simple and self-contained to be worth posting.

Fix an arbitrary interval $[a, b]$, and let $f_{0}$ be an arbitrary non-constant affine function on $[a, b]$.

Inductively, if $f_{n}$ is a piecewise-affine function on $[a, b]$ (not constant on every interval), obtain $f_{n+1}$ by performing the following construction on each affine segment of the graph:

Let $(a_{1}, b_{1})$ and $(a_{2}, b_{2})$ be arbitrary points not lying on a horizontal or vertical line. Replace the (black) segment from $(a_{1}, b_{1})$ to $(a_{2}, b_{2})$ with the (blue) three-segment zig-zag path having vertices $$(a_{1}, b_{1}),\qquad \bigl(\tfrac{1}{3}(a_{1} + 2a_{2}), \tfrac{1}{3}(2b_{1} + b_{2})\bigr),\qquad \bigl(\tfrac{1}{3}(2a_{1} + a_{2}), \tfrac{1}{3}(b_{1} + 2b_{2})\bigr),\qquad (a_{2}, b_{2}).$$

It's easy to see:

• The sequence $(f_{n})$ converges uniformly to a continuous function $f$ on $[a, b]$.

• If $n$ and $k$ are non-negative integers, the functions $f_{n}$ and $f_{n+k}$ (and therefore $f_{n}$ and $f$) agree at every "$3$-adic point", i.e., every point of the form $a + \frac{m}{3^{n}}(b - a)$ with $0 \leq m \leq 3^{n}$. (Geometrically, the recursive construction does not change the values at the endpoints of each "affine segment".)

• For each $n \geq 0$, and for $0 \leq m < 3^{n}$, the function $f_{n+1}$ (and therefore $f$ itself) is not monotone on $\bigl[a + \frac{m}{3^{n}}(b - a), a + \frac{m+1}{3^{n}}(b - a)\bigr]$.

Every open subinterval of $[a, b]$ contains some interval of this form.