Solve $f'=f$ using fourier transform.

Question

I'm trying to solve $f'=f$ using fourier transform. I know that $$\hat f'(\alpha )=2i\pi \alpha \hat f(\alpha )$$ therefore $$f'=f\iff 2i\pi \alpha \hat f(\alpha )=\hat f(\alpha )\iff \hat f(\alpha )=0\iff f(x)=0,$$

what wrong ? Because the solution should be $f(x)=e^{x}$.

Answer 1

The Fourier transform is not defined for the exponential (the integral does not converge). So what you are proving is that $f'=f$ only has the zero solution within the domain of the Fourier transform (say, $L^1(\mathbb R)$).

Note that you say "the solution should be $e^x$". But $f=0\ $ is a solution of $f'=f$, and it is the only one in the domain of the Fourier transform.

Answer 2

$2i\pi \alpha f(\alpha) = f(\alpha)$ implies $f=0$ or $ f$ has support in $\{\alpha = 0\}$. Only for the first case a solution in $L^2$ exists, in the second case it's in the distributional sense.