Exercise 6.4.8

We can use the inequality $\left|\sin <!--\; FUNCTION\; APPLICATION\; -->x\right|\le \left|x\right|$ to show that $f(x)$ converges uniformly by the Weierstrass M-Test over any interval $(-a,a)$, with $M_n=a∕k^2$. Therefore $f$ is defined and is continuous over all real numbers.

The derivative of each term is

which can easily be shown to converge uniformly; hence by the Differentiable Limit Theorem we have that $f(x)$ is differentiable as well. A similar argument shows that $f$ is also twice-differentiable.