Fourier transform of even/odd function How can I show that the Fourier transform of an even integrable function $f\colon \mathbb{R}\to\mathbb{R}$ is even real-valued function? And the Fourier transform of an odd integrable function $f\colon \mathbb{R}\to\mathbb{R}$ is odd and purely imaginary function?
Let $f: \mathbb{R} \to \mathbb{R}$ be an integrable function and let $\hat{f}$ denote its Fourier transform, i.e.
$$
\hat{f}(\xi)=\int_\mathbb{R}e^{ix\xi}f(x)dx.
$$
We have
$$
\overline{\hat{f}(\xi)}=\hat{f}(-\xi)=\int_\mathbb{R}e^{-ix\xi}f(x)dx=\int_\mathbb{R}e^{iy\xi}f(-y)dy.
$$
If $f$ is even then
$$
\overline{\hat{f}(\xi)}=\hat{f}(-\xi)=\int_\mathbb{R}e^{iy\xi}f(y)dy=\hat{f}(\xi),
$$
i.e. $\hat{f}$ is an even real-valued function.

If $f$ is odd then
$$
\overline{\hat{f}(\xi)}=\hat{f}(-\xi)=\int_\mathbb{R}-e^{iy\xi}f(y)dy=-\hat{f}(\xi),
$$
i.e. $\hat{f}$ is an odd purely imaginary function.
The FT of even functions are also even; The FT of odd functions are also odd.
The real part of the FT of a real function is even; The imaginary part of the FT of a real function is odd.


So the Fourier Transform $F(\omega)$ of a real and even function $f(x)$ must satisfy both:


Because $f(x)$ is even: $F(\omega)$ is even (for both real and imaginary parts)
Because $f(x)$ is real: the real part of $F(\omega)$ is even, and the imaginary part is odd


Now for the imaginary part of $F(\omega)$ to be both even and odd, it must be zero, thus $F(\omega)$ is real-only.