So I was trying to prove some properties or rather lemma related to Fourier Transform. I got stuck at this point: I have a function function $$f \enspace \in L^{2}(\mathbb{R} )$$ and its Fourier transform is given by $$\hat{f} $$ The real and imaginary part of the function are indicated respectively by $$Re(f)\enspace and \enspace Im(f)$$ So now I need to prove that : $$If \enspace f\enspace is\enspace real, \enspace then\enspace Re(\hat{f})\enspace is \enspace even\enspace and\enspace the \enspace Im(\hat{f})\enspace is\enspace odd.$$ How can I prove it? Any help.



The Fourier transform of $f$ follows the relation

$$ f(x) = \int_{-\infty}^{+\infty}{\rm d}\xi\;e^{2\pi i \xi x}\hat{f}(\xi) $$

Taking the complex conjugate in both sides

\begin{eqnarray} f^*(x) &=& \int_{-\infty}^{+\infty}{\rm d}\xi\;e^{-2\pi i\xi x}\hat{f}^*(\xi) \stackrel{k=-\xi}{=} -\int_{+\infty}^{-\infty}{\rm d}k\;e^{2\pi ik x}\hat{f}^*(-k) \\ &=& \int_{-\infty}^{+\infty}{\rm d}k\;e^{2\pi ik x}\hat{f}^*(-k) \\ &\stackrel{f\;{\rm is\; real}}{=}& f(x) = \int_{-\infty}^{+\infty}{\rm d}k\;e^{2\pi ik x}\hat{f}(k) \end{eqnarray}

This is true only if

$$ \hat{f}^*(-k) = \hat{f}(k) $$

Or equivalently

\begin{eqnarray} \Re\{\hat{f}(-k)\} -i \Im\{\hat{f}(-k)\} &=& \Re\{\hat{f}(k)\} +i \Im\{\hat{f}(k)\} \\ \Re\{\hat{f}(-k)\} &=& \Re\{\hat{f}(k)\} \quad\mbox{Real part is even}\\ -\Im\{\hat{f}(-k)\} &=& \Im\{\hat{f}(k)\} \quad\mbox{Imaginary part is odd} \end{eqnarray}


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