Fourier transform of even/odd function

Question

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?

Answer 1

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.

Answer 2

Define $F(p) = \int_{-L}^{L} f(x)e^{ipx} dx$. Note that: (let $u=-x$ so $x=-u$ and $dx=-du$ etc..)

$$ F(-p) = \int_{-L}^{L} f(x)e^{-ipx} dx = \int_{L}^{-L} f(-u)e^{ipu}(-du)=\int_{-L}^{L} f(-u)e^{ipu}du$$

Clearly $f(-x) = \pm f(x)$ implies $F(-p) = \pm F(p)$.

Now we turn to the reality part of the claim. Recall $e^{ipx} = \cos px+i\sin px$. Also, remark sine is an odd function whereas cosine is an even function. We know from elementary calculus that the integral of an odd function on $[-L,L]$ vanishes.

1. When $f$ is even then $f(x)\cos(px)$ is even and $f(x)\sin(px)$ is odd. It follows that the imaginary part of Fourier transform vanishes. Consequently, $F(p)$ is real.

2. When $f$ is odd then $f(x)\cos(px)$ is odd and $f(x)\sin(px)$ is even. It follows that the real part of Fourier transform vanishes. Consequently, $F(p)$ is imaginary.

Answer 3

1. The FT of even functions are also even; The FT of odd functions are also odd.
2. 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:

1. Because $f(x)$ is even: $F(\omega)$ is even (for both real and imaginary parts)
2. 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.