let $u(x)$ be a function in $C^2(\mathbb R)$.
Define $u_n$ by
\begin{align*} u_n(x) = u\left(\frac{x}{n}\right) + u\left( - \frac{x}{n}\right) - 2u(0) \end{align*}
then,
\begin{align*} \sum_{n=1}^N u_n \end{align*}
uniformly converges on $[-a, a]$, where $a > 0$.
I know the definition of uniform convergence. And, I just tried transforming $\sum_{n=1}^N u_n $ to $\sum_{n=1}^N \frac{1}{n^2} n^2u_n$ , because $n^2 u_n(x) \rightarrow u^{\prime \prime } (0)$. However I have no idea what to do next.
Hint: Use MVT twice to show that $|u_n(x)|=|[u(\frac x n)-u(0)]+[u(-\frac x n)-u(0)]|$ is bounded by a constant times $\frac 1{n^{2}}$. [You will need the fact that the continuous function $u''$ is bounded on $[-a,a]$]. It now follows by M-test that $\sum u_n$ is uniformly convergent.