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.