My lecture notes have a sort of an open ended question in the sense that the question is:
Let $f:\mathbb{R}\to \mathbb{C}$ and $\hat{f}$ be its Fourier transform. If $f$ is real valued what can we say about $\hat{f}$? On the other hand, if we were to know that $\hat{f}$ is real valued, what do we know about $f$?
If $f$ is real valued, then $\overline{\hat{f}(\xi)} = \hat{f}(-\xi)$ as a quick computation will show. I am suspecting that if $\hat{f}$ is real valued, then $f$ must be an even real valued function. My "rationale" is that $\overline{\hat{f}(\xi)} = \hat{f}(\xi)\Longleftrightarrow \int_{\mathbb{R}}\overline{f(t)}\overline{\exp(-2\pi itv)}dt = \int_{\mathbb{R}}f(t)\exp(-2\pi itv)dt$
whence if $\overline{f(t)}\in \mathbb{R}$ and $f(t) = f(-t)$ then we can show the equality quite easily. But I don't know how to even begin showing this more rigorously. Hence my question is:
Is it true that if the Fourier transform of a function is real valued, then the original function is an even real valued function?
Best Answer
Let's look at $\overline{f(-t)}$:
$ \overline{f(-t)}=\overline{\int_{ - \infty }^{ \infty } { F(\omega) e ^ { j \omega ( - t )} } d \omega} =\int_{ - \infty }^{ \infty } \overline{{ F(\omega) e ^ { - j \omega t } }} d \omega =\int_{ - \infty }^{ \infty } { F(\omega) e ^ { j \omega t } } d \omega=f(t) $
$f$ doesn't have to be real.