# Are continuous functions monotonic for very small ranges?

So I am wondering, if we have a continuous function $f : A \to B$, does a range $[x, x + h]$ exist for each $x\in A$ , $h = h(x)>0$ so that $f$ is monotonic in that range?

• $f(x)=x \sin \frac{1}{x}$ (with $f(0)=0)$ is continuous at $x=0$, but not monitonic in $[0,h]$ for any $h>0$. Mar 20, 2014 at 10:48
• @Sabyasachi: But your $f$ is monotonic in $[0,h]$ for all $h>0$. Mar 20, 2014 at 10:48
• Weierstrass function (en.wikipedia.org/wiki/Weierstrass_function) ? Mar 20, 2014 at 10:59
• @Sabyasachi: You can't do that! The OP said $h \to 0$, so $c$ is presumably fixed. The question was a sensible one, and your interpretation makes it silly. Mar 20, 2014 at 11:02
• @TonyK ah I made some other assumptions. Fair enough, under my interpretation it is silly.
– Guy
Mar 20, 2014 at 11:08

No it is not true for example consider the function $x\sin(\frac{1}{x})$. It is continuous at zero if you take the limit to be the value of the function. But it oscillates very rapidly in every small neighbourhood around zero.

If I am not wrong even everywhere continuous but no where differentiable function has this property.I am referring to the function here

• Yes, any function monotone on an interval is differentiable on a set of positive measure.
– Seth
Mar 20, 2014 at 14:06
• @Seth why is the statement you have stated true? can you please give some justification Mar 20, 2014 at 15:25
• See my answer for an explanation.
– Seth
Mar 20, 2014 at 18:57

And for an example that fails at every $c$, take the Weierstrass fuction defined by $$f(x)=\sum \limits_{n=0}^\infty\left(\dfrac 1 {2^n} \cos \left(15^n \pi x \right)\right).$$

The plot of this function exhibits self-similarity (see red circle below) and looks something like this:

• And a proof or reference it fails the monotonicity condition? Mar 20, 2014 at 11:06
• Nice example. Similar example would be the sample paths of the Wiener process. Mar 20, 2014 at 11:17
• @user2345215: The function can't be monotone on any interval because of the standard result (usually seen in a first course in Lebesgue integration) that a function monotone on an interval $I$ must be (finitely) differentiable at almost every point in $I$ (in the sense of Lebesgue measure), and the fact that "almost everywhere in $I$" implies "dense in $I$" (indeed, much more than this). In other words, a very weak consequence of this is that any function that is monotone in an interval must have at least one point of (finite) differentiability in that interval. Mar 20, 2014 at 13:54
• See this google search in general. More specifically, see: [1] The "Monotone Differentiation Theorem" in these notes by Terry Tao; [2] This 1963 paper; [3] page 100 of these notes (looks like Royden's book). [4] These notes. Mar 20, 2014 at 18:21
• @Cruncher: [I didn't have enough characters left in my last comment to "ping you".] Also, I wanted to add in my last comment that the result is not all that easy to prove, but it's a standard result in virtually every introductory graduate level real analysis course. I do not know if the much weaker result that, for a continuous monotone function and for any interval, there exists at least one point of differentiability. I think it would be interesting if this much weaker version could be proved without appealing to the stronger measure theory version. Mar 20, 2014 at 18:32

Monotone functions are differentiable almost everywhere. So if a function is monotonic on an interval it is differentiable on a set of positive measure. However there are continuous nowhere differentiable functions so it isn't true that continuous functions are monotonic in a (one sided) neighborhood of every point.

An example of a continuous nowhere differentiable function is the Takagi function.

Yet another example: the 1-dimensional Brownian Motion (also called "Wiener Process") is continuous but almost surely, it is not monotonic in any interval.

Proof. let $0\le a\le b$ and denote by $P(a,b)$ the probability that the Brownian motion $B(t)$ is monotonic on the interval $(a,b)$. Then by independence of increments (an important feature of Brownian motion), for each $n$ we have that $$P(a,b) \le 2^{-n} P(a, a+(b-a)2^{-n})$$ hence $P(a,b)=0$.

Since the set of all intervals with rational endpoints is countable, it follows that almost surely $B(t)$ is not monotonic on any such interval. By the density of $\mathbb{Q}$ in $\mathbb{R}$ it follows that any interval contains an interval with rational endpoints, hence with high probability $B(t)$ is not monotonic on any interval.

• I'm not sure this qualifies as an example (as you are not pointing to any one function), but in a sense it is saying that examples exist (in abundance). Indeed most functions that are merely required to be continuous behave like this; it is not pathological but typical behaviour for continuous functions. It is quite remarkable that analysis has long had such an emphasis on differentiable functions, while the awareness that many real-world phenomena are better modelled by such more general continuous functions. Mar 20, 2014 at 13:42
• Brownian motion is a curve, not (necessarily and in fact almost surely not) the graph of a function. Mar 20, 2014 at 21:19
• @IttayWeiss: It's a function of time. Or a probability space which is a function of time. Mar 20, 2014 at 22:37
• @BenVoigt yes, indeed. It is not a function $\mathbb R \to \mathbb R$ (or from a subset of the reals). Mar 20, 2014 at 22:43
• @IttayWeiss: Any given trajectory is $B(t) : \mathbb{R} \rightarrow \mathbb{R}$. The domain is time, which is a real number, and the result is position, also a real number. Mar 20, 2014 at 22:47

Here's an explicit, elementary construction of a continuous but nowhere monotone function, without any appeal to differentiability:

The construction proceeds in phases, with the invariant that after phase $k$, we have chosen finitely many points of the function's graph, such that

1. The horizontal distance between two neighboring points is at most $2^{-k}$, and

2. The slope of the straight line between neighboring points is $\pm\frac{k}{k+1}$, with the sign alternating.

In phase $0$, select the points $(0,0)$ and $(1,0)$

In phase $k\ge 1$, insert two new points between every two neighboring known points $p$ and $q$. Suppose the slope from $p$ to $q$ is positive (namely $\frac{k-1}{k}$); the negative case is the same with opposite signs. Now draw parallel lines of slope $\frac{k}{k+1}$ through $p$ and $q$, and intersect them with the line of slope $-\frac{k}{k+1}$ through the midpoint $\frac12(p+q)$. The two intersections are our two new points. Since $\frac{k}{k+1}>\frac{k-1}{k}$, the two new points will appear in the right order.

After $\omega$ phases, we have selected values for our function at a dense set of $x$ values in $[0,1]$, and restricted to this set the function is uniformly continuous, due to the slope invariant. Therefore it can be uniquely extended to a continuous function $[0,1]\to\mathbb R$.

The extended function is non-monotonic in every open interval. Namely, after finitely many of the phases, at least three points in the interval will have been chosen. If $a,b,c$ are such three consecutive points, $f(b)$ will either be larger than both of $f(a)$ and $f(c)$ or smaller than both of them; in either case $f$ is not monotone in the interval.