We have from the Inverse Fourier Transform
\begin{equation}
g(x)=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}G(k)e^{+ikx}\,dk \tag{6.1.10}
\end{equation}
taking complex conjugate of both sides of equation (6.1.10), we get -
\begin{equation}
g^{*}(x)=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}G^{*}(k)e^{-ikx}\,dk \tag{6.1.11}
\end{equation}
but,
\begin{equation*}
\int\limits_{-\infty}^{\infty}f(x)g^{*}(x)\,dx =\int\limits_{-\infty}^{\infty}f(x)\left\{\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}G^{*}(k)e^{-ikx}\,dk\right\}\,dx
\end{equation*}
\begin{equation*}
=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}f(x)G^{*}(k)e^{-ikx}\,dx\,dk
\end{equation*}
(on changing their order of integration)
\begin{equation*}
= \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}G^{*}(k)\,dk \left[\int\limits_{-\infty}^{\infty}f(x)e^{-ikx}\,dx\right]
\end{equation*}
\begin{equation*}
= \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\mathcal{G}^{*}(k)\,dk \{F(k)\}
\end{equation*}
\begin{equation*}
= \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}F(k)G^{*}(k)\,dk.
\end{equation*}
Again, put \(g(x) = f(x)\) in the above expression, we get -
\begin{equation*}
\int\limits_{-\infty}^{\infty}f(x)f^{*}(x)\,dx = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}F(k)F^{*}(k)\,dk
\end{equation*}
or,
\begin{equation*}
\int\limits_{-\infty}^{\infty}|f(x)|^{2}\,dx = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}|F(k)|^{2}\,dk
\end{equation*}
Note: Similarly, we can prove the following results:
- \begin{equation*} \frac{2}{\pi}\int\limits_{0}^{\infty}F_{s}(k)\mathcal{G}_{s}(k)\,dk = \int\limits_{0}^{\infty}f(x)g(x)\,dx \end{equation*}
- \begin{equation*} \frac{2}{\pi}\int\limits_{0}^{\infty}|F_{s}(k)|^{2}\,dk = \int\limits_{0}^{\infty}|f(x)|^{2}\,dx \end{equation*}
- \begin{equation*} \frac{2}{\pi}\int\limits_{0}^{\infty}F_{c}(k)\mathcal{G}_{c}(k)\,dk = \int\limits_{0}^{\infty}f(x)g(x)\,dx \end{equation*}
- \begin{equation*} \frac{2}{\pi}\int\limits_{0}^{\infty}|F_{c}(k)|^{2}\,dk = \int\limits_{0}^{\infty}|f(x)|^{2}\,dx \end{equation*}
