0
$\begingroup$

In this question the OP tries to prove that only a real, even function $f(t)$ has a real Fourier transform $F(\omega)$.

Is it true? Is this the only case when the Fourier transform is real or are there other $f(t)$ that give rise to a real $F(\omega)$?

  • 1
    Expand $e^{j\omega}$ in the equation of Fourier transform and then set the imaginary part to zero. Then $$\int_{-\infty}^{\infty}f(x)\sin\omega x\,dx=0$$ I doubt that this would necessarily imply that $f(x)$ is even – 2017-01-26
  • 0
    A purely imaginary odd function also has a real Fourier transform. The Fourier transform of $f$ is real if and only if after redefining $f$ on a null set the real part of $f$ is even and the imaginary part of $f$ is odd. – 2017-01-26
  • 0
    @DanielFischer Sorry, what do you mean by "redefining $f$ on a null set"? – 2017-01-26
  • 1
    If two functions differ only on a null set, they have the same Fourier transform. So if we take for example $$f(x) = \begin{cases} i x^7 &, x \in \mathbb{Q} \cap [0,1]\\ \exp(-x^2/2) &, x \notin \mathbb{Q}\cap [0,1]\end{cases}$$ then $f$ has a real even Fourier transform, though $f$ does not not satisfy the conditions. But it's only the null set $\mathbb{Q}\cap [0,1]$ that spoils the fun, and "modulo the values on some null set", $f$ satisfies the conditions. – 2017-01-26

1 Answers 1

1

As Daniel Fischer said, the statement is false. The Fourier transform of $f$ is real-valued if and only if $f(-x) = \overline{f(x)}$ for almost all $x\in\mathbb R$.

To prove the necessity, apply the inverse Fourier transform to a real function $g(\xi)$: ignoring normalization constants, the computation goes as $$ \check g(-x) = \int e^{i(-x)\xi }g(\xi)\,d\xi = \int \overline{e^{ix\xi }}g(\xi)\,d\xi = \overline{\int e^{ix\xi } g(\xi)\,d\xi } = \overline{\check g(x)} $$

The proof of sufficiency is similar: if $f(-x) = \overline{f(x)}$ holds, then $$ \overline{\hat f(x)} = \overline{\int e^{-ix\xi} f(x)\,dx} =\int e^{ix\xi} f(-x)\,dx = \int e^{-ix\xi} f(x)\,dx = \hat f(x) $$ where the second-to-last step is the chang of variable, $x\mapsto -x$.