$\renewcommand{\Re}{\operatorname{Re}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$
In Lecture 15 we introduced Fourier transform in the heuristical way but now we wont a bit more rigorous analysis. Consider smooth fast decaying function $f(x)$ and define its Fourier transform $\hat{f}(k)$ by L5.md:(15.FT) \begin{equation} \hat{f}(\omega)= \frac{1}{2\pi}\int_{-\infty}^\infty f(y)e^{-i\omega y}\,dy, \label{equ-AN.1} \end{equation} and plug it in \begin{equation} g(x)= \int_{-\infty}^\infty \hat{f}(\omega)e^{i\omega x}\,d\omega \label{equ-AN.2} \end{equation} which we consider first as "vp" integral, i.e. as \begin{equation} \lim_{N\to \infty} \int_{-N}^N \hat{f}(\omega)e^{i\omega x}\,d\omega. \label{equ-AN.3} \end{equation} We observe that as $f(x)$ is smooth and fast-decayingnot only (\ref{equ-AN.1}) converges but it could be differentiated (indeed differentiation brings factor $-iy$ under integral but it is covered by assumption of "fast decay") and also decays fast as $\omega\to \infty$ (indeed, integrating by parts we get $i\omega^{-1}f'(y)$ instead of $f(y)$, and we can repeat this procedure.) It is easy to justify completely rigorously that it is sufficient to consider (\ref{equ-AN.3}) and to prove that $g(x)=f(x)$.
But then we get \begin{equation*} \frac{1}{2\pi}\lim_{N\to\infty}\int_{-N}^N \Bigl(\int_{-\infty}^\infty f(y)e^{i\omega (x-y)}\,dy\Bigr)\,d\omega = \frac{1}{2\pi}\lim_{N\to\infty}\int_{-\infty}^\infty f(y)\Bigl(\int_{-N}^N e^{i\omega (x-y)}\,d\omega\Bigr)\,dy \end{equation*} where change of the order of integration could be easily justified; an inner integral in the right-hand expression equals to $2(x-y)^{-1}\sin (N(x-y))$; so we arrive to \begin{equation} \frac{1}{\pi} \lim_{N\to\infty}\int_{-\infty}^\infty f(y) (x-y)^{-1}\sin (N(x-y)) \,dy \label{equ-AN.4} \end{equation}
According to Lemma 13.1 this limit is $0$ provided either $f(x)=0$, or if we integrate over any interval which excludes vicinity of $x$, or if we replace $f(y)$ by $f(y)-f(x)$, in the latter case original (\ref{equ-AN.4}) equals \begin{equation} \frac{1}{\pi} \lim_{N\to\infty}\int_{x-\epsilon}^{x+\epsilon} f(x) (x-y)^{-1}\sin (N(x-y)) \,dy \label{equ-AN.5} \end{equation} where we can select $\epsilon>0$ arbitrarily; but then we can replace $(x-y)^{-1}$ by $(\sin (x-y))^{-1}$ (indeed, their difference vanishes at $x$) but we know from Lecture 13 that then we get $f(x)$.
So inverse Fourier transform of $\hat{f}(\omega)$ indeed restores $f(x)$. Similarly Fourier transform of inverse Fourier transform restores $g(\omega)$ since the difference is only in the signs in front of $i$ and where factor $\frac{1}{2\pi}$ is placed.
Finally, consider $(\hat{f},\hat{g})=\int_{-\infty}^\infty \hat{f}(\omega)\bar{\hat{g}}(\omega)\,d\omega$. Considering again integral in the "vp" sense and plugging $\hat{f}(\omega)$, $\hat{g}(\omega)$ expressed as inverse Fourier transforms we get \begin{multline*} \frac{1}{4\pi^2}\lim_{N\to\infty}\int_{-N}^N \Bigl(\iint_{-\infty}^\infty f(x)\bar{g}(y) e^{i\omega (x-y)}\,dx dy\Bigr)\,d\omega = \\ \frac{1}{2\pi^2} \lim_{N\to\infty}\iint_{-\infty}^\infty (x-y)^{-1}\sin (N(x-y))f(x)\bar{g}(y) \,dx dy \end{multline*} where we again chaned the order of integration and calculated integral by $\omega$. Considering this integral as repeated and applying the same arguments as before to the inner integral one can prove easily that this limit is equal to $\frac{1}{2\pi}\int_{-\infty}^\infty f(x)\bar{g}(x)\,dx= \frac{1}{2\pi}(f,g)$.