Define a function $f: \mathbb{R} \rightarrow \mathbb{R}$ by

$$f(x)=\sum_{n=1}^{\infty} 2^{-n}\left\|2^{n} x\right\|$$

where $\|t\|$ is the distance from $t$ to the nearest integer. Prove that $f$ is continuous. [Results about uniform convergence may not be used unless they are clearly stated and proved.]

Suppose now that $g: \mathbb{R} \rightarrow \mathbb{R}$ is a function which is differentiable at some point $x$, and let $\left(u_{n}\right)_{n=1}^{\infty},\left(v_{n}\right)_{n=1}^{\infty}$ be two sequences of real numbers with $u_{n} \leqslant x \leqslant v_{n}$ for all $n$, $u_{n} \neq v_{n}$ and $u_{n}, v_{n} \rightarrow x$ as $n \rightarrow \infty$. Prove that

$$\lim _{n \rightarrow \infty} \frac{g\left(v_{n}\right)-g\left(u_{n}\right)}{v_{n}-u_{n}}$$

exists.

By considering appropriate sequences of rationals with denominator $2^{-n}$, or otherwise, show that $f$ is nowhere differentiable.