---
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:

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


{% include proof/start.html id="proof-fourier" -%}
We insert the inverse FT into the definition of the inner product:

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

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:

$$\begin{aligned}
    \inprod{\tilde{f}}{\tilde{g}}
    &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}\{f(x)\}\big)^* \: \hat{\mathcal{F}}\{g(x)\} \dd{k}
    \\
    &= A^2 \int
    \Big( \int f^*(x') \: e^{- i s k x'} \dd{x'} \Big)
    \Big( \int g(x) \: e^{i s k x} \dd{x} \Big)
    \dd{k}
    \\
    &= 2 \pi A^2 \iint f^*(x') \: g(x) \Big( \frac{1}{2 \pi}
    \int_{-\infty}^\infty e^{i s k (x - x')} \dd{k} \Big) \dd{x'} \dd{x}
    \\
    &= 2 \pi A^2 \iint f^*(x') \: g(x) \: \delta\big(s (x \!-\! x')\big) \dd{x'} \dd{x}
    \\
    &= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
    = \frac{2 \pi A^2}{|s|} \inprod{f}{g}
\end{aligned}$$
{% include proof/end.html id="proof-fourier" %}


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.