Let $g(x)$ be a real valued differentiable function on $\mathbb{R}$ with $g(0)=0$, and $|g'(x)|\leq M$ for all $x \in \mathbb{R}$. Let $$\displaystyle f(x) = \sum_{n=1}^{\infty} \frac{1}{n} g\bigg(\frac{x}{n}\bigg).$$
(1) Show that $f(x)$ converges for all $x \in \mathbb{R}$, and $f(x)$ is continuous on $\mathbb{R}$.
(2) Is $f(x)$ differentiable on $\mathbb{R}$?
A standard result is that if $f_n:[a,b]\to\mathbb{R}$ are a sequence of differentiable functions and (i) $f_n' \to g$ uniformly and (ii) there is some $x_0$ such that $f(x_0)$ converges, then the $f_n$ converge uniformly to a differentiable $f$ such that $f' = g$. (See Uniform convergence of derivatives, Tao 14.2.7., for example.)
Let $f_n(x) = \sum_{k=1}^{n} {1 \over k} g({x \over k})$. We see that the $f_n$ are differentiable and $f_n(0) = 0$ hence $f_n(0)$ converges.
We note that $f_n'(x) = \sum_{k=1}^{n} {1 \over k^2} g'({x \over k})$ Since $|f_n'(x)| \le \sum_{k=1}^{n} {1 \over k^2} M$, the Weierstraß M-test shows that the $f_n'$ converge uniformly to some $g$. Applying the above result shows that $f_n$ converges to some differentiable $f$ such that $f' = g$.
We see that $f(x) = \sum_{k=1}^{\infty} {1 \over k} g({x \over k})$.
Since $f$ is differentiable, it is continuous. This answers both (i) & (ii).
