Definition: An operator $U\colon H\to H$ is called unitary if $\langle Ux,Uy \rangle=\langle x,y \rangle$ for all $x,y\in H$ and $\text{Im}U=H$.
Currently I'm self studying functional analysis, namely unitary operators. In the text, the author gives the following statement without any proof:
The Fourier transform on $L^2(\mathbb{R})$ $$ (Ff)(\tau):=\frac{1}{\sqrt{2\pi}}\lim_{N\to\infty}\int_{-N}^{N} f(t)e^{it\tau}dt\tag{1} $$ is a unitary operator. The limit is understood in the $L^2(\mathbb{R})$ sense. Moreover, since $F^*=F^{-1}$, the formula for $F^{-1}$ is obtained by changing $i$ to $-i$ in $(1)$.
I know that if $F$ is unitary then $F^{-1}=F^*$. Also, I know that $$ (F^*f)(\tau):= \frac{1}{\sqrt{2\pi}}\lim_{N\to\infty}\int_{-N}^{N} f(t)\overline{e^{it\tau}}dt= \frac{1}{\sqrt{2\pi}}\lim_{N\to\infty}\int_{-N}^{N} f(t)e^{-it\tau}dt,\tag{2} $$ hence I understand since $F^*=F^{-1}$, the formula for $F^{-1}$ is obtained by changing $i$ to $-i$ in $(1)$. What's not obvious to me is that $\langle Ff,Fg \rangle=\langle f,g \rangle$. My attempt at this is: $$ \langle Ff,Fg \rangle=\int_\mathbb{R}(Ff)(\tau)\overline{(Fg)(\tau)}d\tau, $$ and then I've tried expanding this out, etc., but always seem to get nowhere.
