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---
title: "Parseval's theorem"
firstLetter: "P"
publishDate: 2021-02-22
categories:
- Mathematics
- Physics
date: 2021-02-22T21:36:44+01:00
draft: false
markup: pandoc
---
# Parseval's theorem
**Parseval's theorem** is a relation between the inner product of two functions $f(x)$ and $g(x)$,
and 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 FT's definition:
$$\begin{aligned}
\boxed{
\begin{aligned}
\braket{f(x)}{g(x)} &= \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}(k)}{\tilde{g}(k)}
\\
\braket*{\tilde{f}(k)}{\tilde{g}(k)} &= \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)}
\end{aligned}
}
\end{aligned}$$
<div class="accordion">
<input type="checkbox" id="proof-fourier"/>
<label for="proof-fourier">Proof</label>
<div class="hidden">
<label for="proof-fourier">Proof.</label>
We insert the inverse FT into the defintion of the inner product:
$$\begin{aligned}
\braket{f}{g}
&= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}\big)^* \: \hat{\mathcal{F}}^{-1}\{\tilde{g}(k)\} \dd{x}
\\
&= B^2 \int
\Big( \int \tilde{f}^*(k_1) \exp(i s k_1 x) \dd{k_1} \Big)
\Big( \int \tilde{g}(k) \exp(- i s k x) \dd{k} \Big)
\dd{x}
\\
&= 2 \pi B^2 \iint \tilde{f}^*(k_1) \tilde{g}(k) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s x (k_1 - k)) \dd{x} \Big) \dd{k_1} \dd{k}
\\
&= 2 \pi B^2 \iint \tilde{f}^*(k_1) \: \tilde{g}(k) \: \delta(s (k_1 - k)) \dd{k_1} \dd{k}
\\
&= \frac{2 \pi B^2}{|s|} \int_{-\infty}^\infty \tilde{f}^*(k) \: \tilde{g}(k) \dd{k}
= \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}}{\tilde{g}}
\end{aligned}$$
Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
Note that we can equally well do this proof in the opposite direction,
which yields an equivalent result:
$$\begin{aligned}
\braket*{\tilde{f}}{\tilde{g}}
&= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}\{f(x)\}\big)^* \: \hat{\mathcal{F}}\{g(x)\} \dd{k}
\\
&= A^2 \int
\Big( \int f^*(x_1) \exp(- i s k x_1) \dd{x_1} \Big)
\Big( \int g(x) \exp(i s k x) \dd{x} \Big)
\dd{k}
\\
&= 2 \pi A^2 \iint f^*(x_1) g(x) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s k (x_1 - x)) \dd{k} \Big) \dd{x_1} \dd{x}
\\
&= 2 \pi A^2 \iint f^*(x_1) \: g(x) \: \delta(s (x_1 - x)) \dd{x_1} \dd{x}
\\
&= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
= \frac{2 \pi A^2}{|s|} \braket{f}{g}
\end{aligned}$$
</div>
</div>
For this reason, physicists like to define the Fourier transform
with $A\!=\!B\!=\!1 / \sqrt{2\pi}$ and $|s|\!=\!1$, because then it nicely
conserves the functions' normalization.
## References
1. O. Bang,
*Applied mathematics for physicists: lecture notes*, 2019,
unpublished.
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