Fourier series analog of a formula in Fourier transform

Question

Every Fourier transform formula that I know of has a corresponding Fourier series analog, except the multiplication formula $$\int_{-\infty}^\infty f(x)\hat{g}(x) dx=\int_{-\infty}^\infty \hat{f}(y)g(y) dy.$$

So my question is whether a Fourier series analog for this formula exists.

Answer 1

Your formula is another way of writing Parseval's theorem, namely that Fourier transform preserves inner products:

$$\int_{-\infty}^{\infty} f \overline{g} = \int_{-\infty}^{\infty} \hat{f}\overline{\hat{g}} .$$

(You should combine this with the identities $\hat{\hat{g}}(x) = g(-x)$ and $\overline{\hat{g}}(y) = \hat{\overline{g}}(-y).$)

There is also a Fourier series version of Parseval's theorem: namely if $f$ (with period 1) has Fourier series $a_n$ and $g$ (with period 1) has Fourier series $b_n$, then $$\int_0^1 f \overline{g} = \sum_{n = -\infty}^{\infty} a_n \overline{b}_n.$$

However, I don't see how to rewrite this in a way analogous to your formula, just because (unlike in the Fourier transform context, where the function $f$ and the Fourier transform $\hat{g}$ are both functions on the real line) a periodic function and its Fourier series are different beasts, and so it's not clear (to me) how one could mix them together in an integral or something similar.

Added: As p.s. observes in their answer, one could argue that the identity in question is weaker than Parseval, since to deduce Parseval from it requires an application of Fourier inversion (the formula $\hat{\hat{g}}(x) = g(-x)$ used above).

Answer 2

Given an orthogonal function system $\phi_{n}(x)$ ($n=1,2,\dots$) with norm

$$\left\vert \left\vert \phi _{n}\right\vert \right\vert =\sqrt{\left( \phi_{n}\cdot \overline{\phi }_{n}\right) }=\displaystyle\sqrt{\displaystyle\int_{a}^{b}\phi _{n}\left( x\right) \,\overline{\phi }_{n}\left( x\right) \;dx}$$

and two functions $f(x)$ e $g(x)$ represented by the Fourier series

$$f(x)\sim\displaystyle\sum_{n\ge 1} c_{n}\phi_{n}(x)$$

$$g(x)\sim\displaystyle\sum_{n\ge 1} d_{n}\phi_{n}(x)$$

with coefficients $c_n,d_n$

$$c_n=\dfrac{\left ( f\cdot \overline{\phi }_{n}\right)}{\left\vert \left\vert \phi _{n}\right\vert \right\vert ^{2}}=\dfrac{\displaystyle\int_{a}^{b}f\left( x\right)\overline{\phi }_{n}\left( x\right) \;dx}{\left\vert \left\vert \phi _{n}\right\vert \right\vert ^{2}}$$

$$d_n=\dfrac{\left ( f\cdot \overline{\phi }_{n}\right)}{\left\vert \left\vert \phi _{n}\right\vert \right\vert ^{2}}=\dfrac{\displaystyle\int_{a}^{b}g\left( x\right)\overline{\phi }_{n}\left( x\right) \;dx}{\left\vert \left\vert \phi _{n}\right\vert \right\vert ^{2}},$$

and if

$$\displaystyle\int_{a}^{b}\left\vert f(x)-\displaystyle\sum_{n=1}^{\infty}c_{n}\phi_{n}(x)\right\vert ^{2}dx=0$$

and

$$\displaystyle\int_{a}^{b}\left\vert g(x)-\displaystyle\sum_{n=1}^{\infty}d_{n}\phi_{n}(x)\right\vert ^{2}dx=0,$$

then the following equality holds

$$\displaystyle\int_{a}^{b}f(x)\overline{g}_{n}(x)\; dx=\displaystyle\sum_{n\ge 1} c_{n}\overline{d}_{n}||\phi_n||^2,$$

a particular case of which, for $g(x)=f(x)$, is

$$\displaystyle\int_{a}^{b}|f(x)|^2\; dx=\displaystyle\sum_{n\ge 1} |c_{n}|^2||\phi_n||^{2}.$$

Answer 3

There is such a formula, but $f$ and $g$ live on different domains: Let $f: t\mapsto f(t)$ be periodic with period $1$ and $g: k\mapsto g(k)$ be a function defined on $\mathbb Z$. Both these functions have Fourier transforms defined by $$\hat f(k)=\int_0^1 f(t) e^{-2\pi i k t} dt, \ \ \ \hat g(t)=\sum_{k=-\infty}^\infty g(k)e^{-2\pi i k t}.$$ Put $h(t):= \overline{\hat g(t)}$; this is a function of period 1. Then $$h(t)= \sum_{k=-\infty}^\infty \overline{g(k)}e^{2\pi i k t},$$ so the Fourier coefficients of $h$ are given by $\hat h(k)=\overline{g(k)}$. Therefore, by Parseval's theorem for Fourier series one has $$\int_0^1 f(t)\hat g(t)dt = \int_0^1 f(t)\overline{h(t)}dt=\sum_{k=-\infty}^\infty \hat f(k)\overline{\hat h(k)}=\sum_{k=-\infty}^\infty \hat f(k) g(k).$$

Answer 4

Recall the adjoint of an operator $F$ on inner product spaces is defined by $\langle Fg,h \rangle=\langle g, F^*h \rangle$. The identity is just an expression of the fact that the Fourier transform is the complex conjugate of its adjoint: $\langle Fg,\bar{h} \rangle=\langle g, F^* \bar{h} \rangle=\langle \bar{F^*} h,\bar{g} \rangle=\langle Fh,\bar{g} \rangle$. (This is a strictly weaker statement than Parseval's equality.) In general, the conjugate adjoint of the Fourier operator from $H_1$ to $H_2$ is the Fourier operator from $H_2$ to $H_1$.