(This is motivated from the following question)
Fact: If $f \in C^1(\mathbb{T})$, then the Fourier coefficients $\widehat{f'}(n)$ of the derivative $f′$ can be expressed in terms of the Fourier coefficients $\hat{f}(n)$ of the function $f$, via the formula $\widehat{f'}(n) = in \hat{f}(n)$
Take $f(\theta)=e^{\theta}, \theta \in (-\pi, \pi).$ Note that $f'(\theta)=f(\theta)$ and $f\in C^{1}(\mathbb T);$ by above fact $\hat{f}(n)=in \hat{f}(n)$ for all $n\in \mathbb Z.$ and so $\hat{f}(n)(1-in)=0$ for all $n\in \mathbb Z.$ Since $\mathbb C$ is an integral domain, $\hat{f}(n)=0$ for all $n\in \mathbb Z.$ But if this the case then $f$ is identically zero function which certainly not true!!!!
So my Question is: Where is the fallacy in the above argument ?
Thanks,
$f(\theta)=e^\theta$ is not $C^1$, not even continuous, on $\mathbb{T}$. It is continuous on $[\,0,2\,\pi\,]$, but $f(0)\ne f(2\,\pi)$.