A function which is continuous everywhere, but not differentiable at any point Let $f: \mathbb{R} \to \mathbb{R}$ be the function 
$$f(x) : = \sum_{n=1}^\infty 4^{-n} \cos (32^n \pi x).$$
(a) Show that this series is uniformly convergent, and that $f$ is continuous. 
(b) Show that for every integer $j$ and every integer $m \ge 1$, we have  
$$\left|f\left(\frac{j+1}{32^m}\right) - f\left(\frac{j}{32^m}\right)\right| \ge 4^{-m}.$$ 
(Hint: use the identity 
$$\sum_{n=1}^\infty a_n = \left(\sum_{n=1}^{m-1} a_n\right) + a_m + \sum_{n=m+1}^\infty a_n$$ 
for certain sequences $a_n$. Also use the fact that the cosine function is periodic with period $2 \pi$, as well as the geometric series formula. Finally you will need the inequality $|\cos(x) - \cos(y)| \le |x -y|$ for any real number $x$ and $y$. 
(c) Using (b), show that for every real number $x_0$, the function $f$ is not differentiable at $x_0$. (Hint: for every $x_0$ and every $m\ge 1$, there exists an integer $j$ such that $j \le 32^mx_0 \le j+1$.)
(a) can be shown using the Weierstrass M test. $4^{-n} \cos(32^n \pi x) \le 4^{-n}$ for all $x$, and $\sum_{n=1}^\infty 4^{-n} < \sum_{n=1}^\infty 2^{-n} =1$. 
For (b), I first expand the equation out as the hint suggests 
$$\sum_{n=1}^{m-1} 4^{-n} [\cos (32^n \pi \frac{j+1}{32^m})) -\cos (32^n \pi \frac{j}{32^m})]  + 4^{-m} [\cos (\pi (j+1)) -\cos (\pi (j))] + \sum_{n=m+1}^\infty 4^{-n} [\cos (32^n \pi \frac{j+1}{32^m})-\cos (32^n \pi \frac{j}{32^m}) ].$$
It turns out that the middle term is either $4^{-m}2$ if $j$ is odd or $4^{-m}(-2)$ if $j$ is even. The last term is cancelled out due to $2\pi$ periodicity. However, I am not sure how to deal with the first term, and why we need the inequality $|\cos(x) - \cos(y)| \le |x -y|$.  
I also appreciate if you give some hint for (c). 
Thanks in advance. 
 A: For (b), to bound $|A+B|$ from below, use $|A+B|\geqslant|B|-|A|$ and bound $|A|$ from above (here, $A$ is your first term, and $B$ is the remaining expression, with $|B|=2\cdot 4^{-m}$ found by you): \begin{align}\big|f\big(32^{-m}(j+1)\big)-f\big(32^{-m}j\big)\big|&\geqslant 2\cdot 4^{-m}-\sum_{n=1}^{m-1}4^{-n}\underbrace{\big|\cos\big(32^{n-m}\pi(j+1)\big)-\cos\big(32^{n-m}\pi j\big)\big|}_{\leqslant\ 32^{n-m}\pi\text{ by the suggested inequality}}\\&\geqslant 2\cdot 4^{-m}-32^{-m}\pi\underbrace{\sum_{n=1}^{m-1}8^n}_{=(8^m-8)/7}>(2-\pi/7)4^{-m}>4^{-m}.\end{align}
For (c), the differentiability of $f$ at $x_0$, i.e. the existence of $L=\lim\limits_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}$ would imply $L=\lim\limits_{m\to\infty}\frac{f(a_m)-f(x_0)}{a_m-x_0}$ for any sequence $\{a_m\}$ with $\lim\limits_{m\to\infty}a_m=x_0$ and $a_m\neq x_0$ for each $m$. Or even $L=\lim\limits_{m\to\infty}\frac{f(b_m)-f(a_m)}{b_m-a_m}$ for any two sequences $\{a_m\}, \{b_m\}$ with $\lim\limits_{m\to\infty}a_m=\lim\limits_{m\to\infty}b_m=x_0$ and $a_m<b_m$ for each $m$. Now let $j_m=\lfloor32^m x_0\rfloor$ as suggested essentially, and consider $a_m=32^{-m}j_m$ and $b_m=32^{-m}(j_m+1)$. Then the conditions are met but, by (b), $$\left|\frac{f(b_m)-f(a_m)}{b_m-a_m}\right|\geqslant\frac{4^{-m}}{32^{-m}}=8^m,$$ hence $\lim\limits_{m\to\infty}\frac{f(b_m)-f(a_m)}{b_m-a_m}$ cannot exist.
