What condition on the coefficients $a_n$ will guarantee $f(x) = \Sigma_{n = -\infty}^{\infty}a_{n}e^{2\pi inx}$ is k times differentiable? $f(x) = \Sigma_{n = -\infty}^{\infty}a_{n}e^{2\pi inx}$
What condition on the coefficients $a_n$ will guarantee $f$ is $k$ times differentiable?
I'm not sure where to begin with this, because it appears to me that if you differentiate each term in the sum individually, the function is $k$ times differentiable for all $k$. What am I missing here?
 A: Let's start with the case $k = 0$. To guarantee that $\sum_{n=-\infty}^{\infty} a_n e^{2\pi i n x}$ will converge to a continuous function, we can try and force the series to converge uniformly and then the limit will be continuous (as a uniform limit of continuous functions). In order to force uniform convergence, we can try and bound the series uniformly and apply the Wierstrass M-test. We have
$$ \sum_{n=-N}^{N} \left| a_n e^{2\pi i n x} \right| = \sum_{n = -N}^{N} |a_n| $$
and so if $\sum_{n = -\infty}^{\infty} |a_n| < \infty$ the series will converge uniformly and will be continuous.
Regarding $k = 1$, we can differentiate the series formally and then require that the formal series of the derivative will converge uniformly. By standard results regarding series of functions, it will imply that the limit is indeed differentiable (even continuously differentiable). Thus, if
$$ \sum_{n=-\infty}^{\infty} \left| 2 \pi i n a_n e^{2\pi i n x} \right| = 2 \pi \sum_{n=\infty}^{\infty} |n a_n| < \infty $$ 
the original series will converge to a continuously differentiable function. You can continue in this way to obtain conditions for $k > 1$.
