Theorem: For $f, g \in L^2(\mathbb{R})$ with Fourier transforms $\tilde{f}, \tilde{g}$:

$\int_{-\infty}^{\infty}f^*(x)g(x)dx = \frac{1}{2\pi}\int_{-\infty}^{\infty}\tilde{f}^*(k)\tilde{g}(k)dk$

Proof:

Step 1: First prove for $f, g \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ (integrable and square-integrable).

Start with the inverse Fourier transform:

$g(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\tilde{g}(k)e^{ikx}dk$

Substitute into the left side:

$\int_{-\infty}^{\infty}f^*(x)g(x)dx = \int_{-\infty}^{\infty}f^*(x)\left[\frac{1}{2\pi}\int_{-\infty}^{\infty}\tilde{g}(k)e^{ikx}dk\right]dx$

Step 2: Interchange the order of integration (justified by Fubini's theorem since $f \in L^2$ and $\tilde{g} \in L^1$):

$= \frac{1}{2\pi}\int_{-\infty}^{\infty}\tilde{g}(k)\left[\int_{-\infty}^{\infty}f^*(x)e^{ikx}dx\right]dk$

The inner integral is:

$\int_{-\infty}^{\infty}f^*(x)e^{ikx}dx = \left[\int_{-\infty}^{\infty}f(x)e^{-ikx}dx\right]^* = \tilde{f}^*(k)$

Step 3: Therefore:

$\int_{-\infty}^{\infty}f^*(x)g(x)dx = \frac{1}{2\pi}\int_{-\infty}^{\infty}\tilde{f}^*(k)\tilde{g}(k)dk$

Step 4: Extend to all $f, g \in L^2(\mathbb{R})$ by density. The space $L^1 \cap L^2$ is dense in $L^2$, so for any $f \in L^2$, there exists a sequence $\{f_n\} \subset L^1 \cap L^2$ with $\|f_n - f\|_2 \to 0$.

By continuity of the inner product:

$\langle f, g\rangle = \lim_{n \to \infty}\langle f_n, g\rangle = \lim_{n \to \infty}\frac{1}{2\pi}\langle\tilde{f}_n, \tilde{g}\rangle$

We must show that $\tilde{f}_n \to \tilde{f}$ in $L^2$ as well. This requires proving that the Fourier transform extends continuously to $L^2$.

Step 5: For $f \in L^1 \cap L^2$, by Step 3 with $g = f$:

$\int_{-\infty}^{\infty}|f(x)|^2dx = \frac{1}{2\pi}\int_{-\infty}^{\infty}|\tilde{f}(k)|^2dk$

This shows the Fourier transform is an isometry on $L^1 \cap L^2$. By completeness of $L^2$ and density of $L^1 \cap L^2$, the Fourier transform extends uniquely to an isometry on all of $L^2(\mathbb{R})$.

Step 6: The general inner product formula follows by polarization identity:

$\langle f, g\rangle = \frac{1}{4}\left[\|f + g\|^2 - \|f - g\|^2 + i\|f + ig\|^2 - i\|f - ig\|^2\right]$

and applying Step 5 to each term.

Theorem: For $f(x)$ with period $2L$ and Fourier coefficients $c_n = \frac{1}{2L}\int_{-L}^L f(x)e^{-in\pi x/L}dx$:

$\frac{1}{2L}\int_{-L}^L |f(x)|^2dx = \sum_{n=-\infty}^{\infty}|c_n|^2$

Proof:

Expand $f(x)$ and $f^*(x)$ in Fourier series:

$f(x) = \sum_{n=-\infty}^{\infty} c_n e^{in\pi x/L}, \quad f^*(x) = \sum_{m=-\infty}^{\infty} c_m^* e^{-im\pi x/L}$

Then:

$|f(x)|^2 = f^*(x)f(x) = \sum_{m,n} c_m^* c_n e^{i(n-m)\pi x/L}$

Integrate over one period:

$\frac{1}{2L}\int_{-L}^L |f(x)|^2dx = \sum_{m,n} c_m^* c_n \left[\frac{1}{2L}\int_{-L}^L e^{i(n-m)\pi x/L}dx\right]$

The integral is:

$\frac{1}{2L}\int_{-L}^L e^{i(n-m)\pi x/L}dx = \begin{cases}1 & n = m \\ 0 & n \neq m\end{cases}$

by orthogonality of exponentials. Thus:

$\frac{1}{2L}\int_{-L}^L |f(x)|^2dx = \sum_{n=-\infty}^{\infty}c_n^* c_n = \sum_{n=-\infty}^{\infty}|c_n|^2$