We prove $\|f\|_{L^2}^2 = \|\hat{f}\|_{L^2}^2$ for $f \in \mathcal{S}(\mathbb{R}^n)$ (Schwartz space), then extend by density to $L^2$.

Step 1: Compute $\|\hat{f}\|_{L^2}^2$

By definition: $\|\hat{f}\|_{L^2}^2 = \int_{\mathbb{R}^n} |\hat{f}(\xi)|^2\,d\xi = \int_{\mathbb{R}^n} \hat{f}(\xi)\overline{\hat{f}(\xi)}\,d\xi$

Step 2: Use Fubini's Theorem

$\overline{\hat{f}(\xi)} = \int_{\mathbb{R}^n} \overline{f(y)}e^{2\pi i y \cdot \xi}\,dy$

Therefore: $\|\hat{f}\|_{L^2}^2 = \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} f(x)\overline{f(y)}e^{-2\pi i (x-y) \cdot \xi}\,dx\,dy\,d\xi$

Step 3: Evaluate the $\xi$ Integral

The inner integral is: $\int_{\mathbb{R}^n} e^{-2\pi i (x-y) \cdot \xi}\,d\xi = \delta(x - y)$

in the sense of distributions. More rigorously, for Schwartz functions: $\int_{\mathbb{R}^n} e^{-2\pi i z \cdot \xi}\,d\xi = \delta(z)$

Step 4: Apply the Delta Function

$\|\hat{f}\|_{L^2}^2 = \int_{\mathbb{R}^n}\int_{\mathbb{R}^n} f(x)\overline{f(y)}\delta(x-y)\,dx\,dy = \int_{\mathbb{R}^n} f(x)\overline{f(x)}\,dx = \|f\|_{L^2}^2$

Step 5: Extension to $L^2$

We've proved the result for $f \in \mathcal{S}$. Since $\mathcal{S}$ is dense in $L^2$ and both $\mathcal{F}$ and the identity map are continuous:

For any $f \in L^2$, choose $f_n \in \mathcal{S}$ with $f_n \to f$ in $L^2$. Then: $\|\hat{f}\|_{L^2}^2 = \lim_{n \to \infty}\|\hat{f_n}\|_{L^2}^2 = \lim_{n \to \infty}\|f_n\|_{L^2}^2 = \|f\|_{L^2}^2$

Step 6: Unitarity

The relation $\|\hat{f}\|_{L^2} = \|f\|_{L^2}$ shows $\mathcal{F}$ is an isometry. Combined with the inversion formula, $\mathcal{F}$ is unitary: $\langle \mathcal{F}f, \mathcal{F}g \rangle = \langle f, g \rangle$

for all $f, g \in L^2$.