Decay of Fourier coefficients sequence If $f:\Bbb R\to \Bbb R$ is a $2\pi-$ periodic, $C^1$ function, then $k^2a_{k}(f)\to 0$ where $$a_{k}(f)=\frac {1}{\pi}\int_{-\pi}^{\pi}f(x)\cos kx dx$$ 
are the Fourier coefficients.
I ask if this is true. It's not an exercise but a thought of mine which I struggle to solve (just started to learn Fourier Analysis). Can you find a counterexample?
 A: No, this is not true.
Let us start by constructing a $2\pi$ periodic function which is $C^1$ everywhere, but $C^2$ nowhere. Define
$$g(x)=\sum_{n=1}^\infty \frac{\sin(100^n x)}{2^n}$$
Since Weierstrass it is known, that this function is everywhere continuous, but nowhere differentiable.
Therefore the function
$$
\begin{align}
f(x)&=-\int_0^x g(t) dt=-\sum_{n=1}^\infty \int_0^x \frac{\sin(100^n t)}{2^n} dt =\sum_{n=1}^\infty \left(\frac{\cos(100^n x)}{200^n}-\frac{1}{200^n}\right)\\
&=-\frac{1}{199}+\sum_{n=1}^\infty \frac{\cos(100^n x)}{200^n}
\end{align}
$$
is $C^1$ everywhere, but nowhere $C^2$, because $f^\prime(x)=-g(x)$ (both holds by the fundamental theorem of calculus). Note that the interchange of sum and integral is justified, because the series converges absolutely and uniformly in $x$.
Now the Fourier coefficients can be conveniently read off as $a_k=\frac{1}{200^n}$ if $k=100^n$, $n\ge 1$. 
Since $(100^{n})^2a_{100^n}=\frac{100^{2n}}{200^n}=50^n\rightarrow\infty$, this is a counterexample to the proposed statement.
Note: 
$100$ is of course not the smallest number such that this works.
