---
title: "Parseval's theorem"
sort_title: "Parseval's theorem"
date: 2021-02-22
categories:
- Mathematics
- Physics
layout: "concept"
---

**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:

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

<div class="accordion">
<input type="checkbox" id="proof-fourier"/>
<label for="proof-fourier">Proof</label>
<div class="hidden" markdown="1">
<label for="proof-fourier">Proof.</label>
We insert the inverse FT into the defintion of the inner product:

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

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:

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

</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.