For any complex-valued $f \in L^1(\mathbb{R})$, let's define its Fourier transform $\hat{f}$ with the following convention
$$
\hat{f}(\omega) := \int_{\mathbb{R}} f(x) e^{-i \omega x} dx
$$
I would like a confirmation of the following:
- $f$ even $\Rightarrow$ $\hat{f}$ even
- $f$ odd $\Rightarrow$ $\hat{f}$ odd
To prove for example the first statement, I would argue that
$$
\hat{f}(\omega)
= \int_{\mathbb{R}} f(x) e^{-i \omega x} dx
= \int_{\mathbb{R}} f(-x) e^{i \omega x} dx
= \int_{\mathbb{R}} f(x) e^{i \omega x} dx
= \int_{\mathbb{R}} f(x) e^{-i (-\omega) x} dx
= \hat{f}(-\omega)
$$
where the second equality is a consequence of the fact that the Lebesgue measure ($\lambda$) satisfies $\lambda(A) = \lambda(-A)$, for every Borel set $A$ and the third is just using the hypothesis, i.e. that $f(x) = f(-x)$.
The second statement can be proved in the very same way.
Is the above correct?!?
I started having second thoguhts after seeing this heavily downvoted answer
Fourier transform of even/odd function
and especially its first comment.
Best Answer
Yes, you are correct. The corresponding parity of the Fourier transform follows from the parity of the function and the fact that the Lebesgue integral is invariant under reflection (which follows from the relative invariance of Lebesgue measure).