Problem: For all $n \in \mathbb{N}$ let $g_n : (-1,1) \to \mathbb{R}$ be a twice differentiable function. Assume that $g_n(0) = g' _n(0) = 0$ for all $n$ and suppose that there exists some $M \in \mathbb{R}$ so that for all $n \in \mathbb{N}$ and all $x \in (-1,1)$ we have $|g_n '' (x) | \leq M$.
Define $g(x) = \begin{cases} g_n(x), & x = \frac{1}{P_n ^m} \text{ for some } m \in \mathbb{N} \\ 0, & \text{otherwise}. \end{cases}$
Prove that $g$ is differentiable at $0$.
I am not sure about how to proceed with this problem or if my approach is correct. I believe that I should prove that for some $\epsilon >0$ there exists $\delta >0$ such that $|x| < \delta$ implies $|g_n ' (x)| < \epsilon$ simultaneously for all $n$. Any suggestions on how to work this problem, or if this approach is completely false? (I am using baby Rudin and reached L'Hôpital's Rule in the chapter concerning differentiability.)
I suppose $p_n$ in your definition stands for the n-th prime number. By MVT $|g'_n (x)| =|g'_n (x)-g'_n(0)| \leq M|x|$. Hence, by another application of MVT $|g_n (x)| =|g_n (x)-g_n(0)| \leq M x^{2}$. Hence $|\frac {g(p_n^{-m}) -g(0)} {p_n^{-m}}| \leq M(p_n^{-m})^{2} / {p_n^{-m}}=M{p_n^{-m}} \leq Mp_n^{-1} \to 0$. Hence $g'(0) =0$.