Answer: The Nowhere-Monotone Continuous Function

Key Idea / Intuition

The answer is yes — such functions exist, and in fact the "generic" continuous function (in the Baire category sense) is nowhere monotone. The key insight is that continuity does not force any local monotone behavior: you can construct a function that oscillates infinitely on every interval, never settling into any increasing or decreasing trend. The Weierstrass nowhere-differentiable function is the canonical example, and we explain why nowhere differentiability (with unbounded variation) forces nowhere monotonicity.

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Formal Proof / Solution

Step 1: The Weierstrass Function Is Continuous

Define the classical Weierstrass function: $f(x) = \sum_{n=0}^{\infty} a^n \cos(b^n \pi x),$ where $0 < a < 1$, $b$ is a positive odd integer, and $ab > 1 + \frac{3\pi}{2}$.

This converges uniformly (by Weierstrass $M$-test, since $\sum a^n < \infty$), so $f$ is continuous.

Step 2: Nowhere Differentiability Implies Nowhere Monotone

Claim: If $f$ is monotone on some interval $(c, d)$, then $f$ is differentiable almost everywhere on $(c, d)$.

Proof of claim: This is Lebesgue's monotone differentiation theorem — any monotone function on an interval is differentiable almost everywhere.

Therefore: if $f$ were monotone on any open interval $(c, d)$, it would have to be differentiable at almost every point of $(c, d)$.

But the Weierstrass function is differentiable nowhere. This is a contradiction.

Hence $f$ is nowhere monotone.

Step 3: Why Is Nowhere Monotone Surprising?

One might think: "Surely on some tiny interval, the function must go up or down overall." But continuity alone does not prevent infinite oscillation. The Weierstrass function oscillates on every scale — zooming into any interval reveals the same chaotic structure. There is no interval where the function "trends" in any direction.

Step 4: Baire Category Perspective (Bonus)

Define the set: $M_n = \left\{ f \in C([0,1]) : \exists\, x \in [0, 1-1/n] \text{ such that } f \text{ is increasing on } [x, x+1/n] \right\}.$

Each $M_n$ is closed and nowhere dense in $C([0,1])$ with the sup-norm. By the Baire Category Theorem, $C([0,1]) \setminus \bigcup_n M_n$ is a dense $G_\delta$ — meaning most continuous functions (in the topological sense) are nowhere monotone.

Summary

$\boxed{\text{Yes. The Weierstrass function is continuous and nowhere monotone.}}$

The logical chain is: $\text{nowhere differentiable} \xRightarrow{\text{Lebesgue}} \text{differentiable a.e. on no interval} \Rightarrow \text{monotone on no interval.}$