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?
 A: 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.
A: 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.
A: 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


*

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

*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.
A: 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
A: 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:

