Are the Fourier transforms of two orthogonal functions orthogonal?

Question

I know that if a matrix $\mathbf{A}$ is orthogonal, then $x\cdot x = \mathbf{A}x\cdot \mathbf{A}x$ because $\mathbf{A}x\cdot \mathbf{A}x = (\mathbf{A}x)^T \mathbf{A}x = x^T\mathbf{A}^T\mathbf{A}x=x^T\mathbf{A}^{-1}\mathbf{A}x=x^Tx=x\cdot x$.

If the inner product over functions is chosen to be, $$ f\cdot g = \int_{-\infty}^{\infty} f(x)\overline{g(x)} \,\mathrm{dx} $$

Then it seems intuitive to me that if I normalize the Fourier transform the right way, then it will act like it's "orthonormal," and preserve this inner product so that $\mathcal{F}[f]\cdot \mathcal{F}[g] = f\cdot g$. How do I prove this, and which normalization should I use to make it true?

Answer 1

You are looking for the Plancherel theorem. A part of it states that the Fourier transform preserves the inner product in $\mathscr{L}^2$.

The result is that, $$ \int_{-\infty}^\infty f(x)\overline{g(x)} \, dx = \int_{-\infty}^\infty \widehat{f}(\xi) \overline{\widehat{g}(\xi)} \, d\xi $$ When both $f$ and $g$ are in $\mathscr{L}^2$.

Answer 2

For a proof define $\hat{g}(\xi)=\int_\mathbb{R}g(x)\exp -2\pi ix\xi dx$ so $g(x)=\int_\mathbb{R}\hat{g}(\xi)\exp 2\pi ix\xi dx$ and both sides of the Plancherel theorem are $\int_{\mathbb{R}^2}f(x)\overline{\hat{g}(\xi)}\exp -2\pi ix\xi dx$.