Prove that $f \in C^k$ if $|\hat{f}(n)|\leq C/|n|^{k+a}$ I'm looking at some problems related to Fourier series. This one stumped me a little.

Suppose that $f$ is $2\pi$-periodic and piecewise smooth. Show that if
  there exist $k \in \mathbb{N}, a > 0,$ and $C > 0$ such that
  $$|\hat{f}(n)|\leq \frac{C}{|n|^{k+a}},$$ for $n \geq N_0$, then $f
 \in C^k$.

Note that $\hat{f}(n)$ is the Fourier coefficients of $f$. I really seem to have no idea where to start with this one. I was thinking about using the expression for the Fourier coefficients $$\hat{f}(n) = \frac{1}{2\pi}\int_0^{2\pi} f(t) \mathrm{e}^{-int}\, \mathrm{d}t$$ and bound it by $|\hat{f}(n)| \leq 1/2\pi \int_0^{2\pi} |f(t)|\, \mathrm{d}t$. But this throws the $n$ out of the question.
Any ideas?
 A: As stated, the proposition doesn't hold. Consider the $2\pi$-periodic extension $f$ of $\chi_{[-\pi/2,\pi/2]}$. We have
$$\begin{align}
\hat{f}(n) &= \frac{1}{2\pi}\int_{-\pi}^\pi f(t) e^{-int}\,dt\\
&= \frac{1}{2\pi} \int_{-\pi/2}^{\pi/2} e^{-int}\,dt\\
&= \frac{i}{2\pi n}\left(e^{-in\pi/2} - e^{in\pi/2}\right)\\
&= \frac{i((-i)^n - i^n)}{2\pi n},
\end{align}$$
which is $O\left(\frac{1}{\lvert n\rvert}\right)$, but $f \notin C^0$.
If we amend the premises to $a > 1$, then an integration by parts shows
$$\begin{align}
\hat{f}(n) &= \frac{1}{2\pi}\int_0^{2\pi} f(t) e^{-int}\,dt\\
&= \frac{1}{2\pi}\left[\frac{i}{n}f(t)e^{-int}\right]_0^{2\pi} + \frac{1}{2\pi i n} \int_0^{2\pi} f'(t)e^{-int}\,dt\\
&= \frac{1}{in} \hat{f'}(n)
\end{align}$$
for differentiable $f$ with nice enough (need not be continuous) derivative.
Now, if
$$\lvert \hat{f}(n)\rvert \leqslant \frac{C}{\lvert n\rvert^{k+a}}$$
for all large enough $n$ and some $k\in\mathbb{N}$ and $a > 1$, then the $k$ times termwise differentiated Fourier series converges uniformly,
$$g_k(t) = \sum_{n\in\mathbb{Z}} \hat{f}(n)(in)^k e^{int}$$
is a continuous function. Integrating $k$ times, and inserting the constant term $\hat{f}(0)$ exposes $f$ as a $k$-times continuously differentiable function.
A: I would write
$$
f(t) = \sum_{n} \hat f (n) e^{int}
$$
and notice that the given condition assures that the series of $k$-th derivatives converges uniformly. This implies that $f$ is differentiable $k$ times.
