% Parseval's theorem


# Parseval's theorem

**Parseval's theorem** relates the inner product of two functions $f(x)$ and $g(x)$ to the
inner product of their [Fourier transforms](/know/concept/fourier-transform/)
$\tilde{f}(k)$ and $\tilde{g}(k)$.
There are two equivalent ways of stating it,
where $A$, $B$, and $s$ are constants from the Fourier transform's definition:

$$\begin{aligned}
    \boxed{
        \braket{f(x)}{g(x)} = \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}(k)}{\tilde{g}(k)}
    }
    \\
    \boxed{
        \braket*{\tilde{f}(k)}{\tilde{g}(k)} = \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)}
    }
\end{aligned}$$

For this reason, physicists like to define their Fourier transform
with $A = B = 1 / \sqrt{2\pi}$ and $|s| = 1$, because then the FT nicely
conserves the total probability (quantum mechanics) or the total energy
(optics).

To prove this, we insert the inverse FT into the inner product
definition:

$$\begin{aligned}
    \braket{f}{g}
    &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}\big)^* \: \hat{\mathcal{F}}^{-1}\{\tilde{g}(k)\} \dd{x}
    \\
    &= B^2 \int
    \Big( \int \tilde{f}^*(k_1) \exp(i s k_1 x) \dd{k_1} \Big)
    \Big( \int \tilde{g}(k) \exp(- i s k x) \dd{k} \Big)
    \dd{x}
    \\
    &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \tilde{g}(k) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s x (k_1 - k)) \dd{x} \Big) \dd{k_1} \dd{k}
    \\
    &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \: \tilde{g}(k) \: \delta(s (k_1 - k)) \dd{k_1} \dd{k}
    \\
    &= \frac{2 \pi B^2}{|s|} \int_{-\infty}^\infty \tilde{f}^*(k) \: \tilde{g}(k) \dd{k}
    = \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}}{\tilde{g}}
\end{aligned}$$

Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
Note that we can just as well do it in the opposite direction,
which yields an equivalent result:

$$\begin{aligned}
    \braket*{\tilde{f}}{\tilde{g}}
    &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}\{f(x)\}\big)^* \: \hat{\mathcal{F}}\{g(x)\} \dd{k}
    \\
    &= A^2 \int
    \Big( \int f^*(x_1) \exp(- i s k x_1) \dd{x_1} \Big)
    \Big( \int g(x) \exp(i s k x) \dd{x} \Big)
    \dd{k}
    \\
    &= 2 \pi A^2 \iint f^*(x_1) g(x) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s k (x_1 - x)) \dd{k} \Big) \dd{x_1} \dd{x}
    \\
    &= 2 \pi A^2 \iint f^*(x_1) \: g(x) \: \delta(s (x_1 - x)) \dd{x_1} \dd{x}
    \\
    &= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
    = \frac{2 \pi A^2}{|s|} \braket{f}{g}
\end{aligned}$$