In a course on Fourier Analysis, the proof of Parseval's Theorem ends on the following line:
$$\left<f,g\right>_{\mathcal{H}} = \left<\mathcal{F}f,\mathcal{F}g\right>_{\ell^2}$$
I can then show that:
$$\left<f,g\right> = \left<f,\mathcal{F}^*\mathcal{F}g\right> \implies \mathcal{F}^*\mathcal{F}g = g \implies \mathcal{F}^*\mathcal{F} = I.$$
I am wondering how this is enough to conclude that $\mathcal{F}$ is unitary, since we also need to verify that $\mathcal{F}\mathcal{F}^* = I$.
At this point, we have not shown that $\mathcal{F}$ is invertible, but if it were I can see how uniqueness of the inverse may be used, since $\mathcal{F}^{-1}\mathcal{F}=I$. Any help would be appreciated!
You have shown that $\mathcal{F}^*\mathcal{F}=I$. Then you can easily show that $\mathcal{G}f = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{ixt}f(t)dt$ also satisfies $\mathcal{G}^*\mathcal{G}=I$. Finally, show that $\mathcal{F}^*=\mathcal{G}$ and you have what you want.