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Let $f,g$ be two continuous functions with period$=1$.

Are the Fourier coefficients of $f*g$ are given by the products $f(n)g(n)$ (of the $n$-th coefficient in each series)?

Thanks!

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    $f$,g:R-->C (must use letters)2012-01-06

2 Answers 2

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The following is the pedestrian version of AD.'s answer:

Put ${\mathbb R}/{\mathbb Z}=: T$ (for one-dimensional torus). Then a function $f\!: T\to {\mathbb C}$ has a Fourier transform $\hat f\!: {\mathbb Z}\to{\mathbb C}$ given by $\hat f(n)=\int_T f(t) \exp(-2n\pi i t)\ {\rm d}t\qquad (n\in{\mathbb Z}) .$ The numbers $\hat f(n)$ are nothing else but the ordinary complex Fourier coefficients of $f$.

Given two such functions $f$ and $g$ there is their convolution $(h:=) \quad f*g\!: \ T\to{\mathbb C}$ which is defined as follows: $h(x):=\int_T f(x-t) g(t)\ {\rm d}t\qquad(x\in T)\ .$ Calculating the complex Fourier coefficients of $h$ amounts to an integral over $T\times T$. Using Fubini's theorem one gets \eqalign{\hat h(n)&=\int_T h(x)\exp(-2n\pi i x)\ {\rm d}x = \int_T\left(\int_T f(x-t)g(t)\ {\rm d}t\right)\ \exp(-2n\pi i x)\ {\rm d}x \cr &=\int_{T\times T} f(x-t) g(t)\exp\bigl(-2n\pi i x \bigr)\ {\rm d}x'\ {\rm d}t \cr &=\int_{T\times T} f(x') g(t)\exp\bigl(-2n\pi i(x'+t)\bigr)\ {\rm d}x'\ {\rm d}t \cr &=\int_T f(x')\exp(-2n\pi i x')\ {\rm d}x' \cdot \int_T g(t)\exp(-2n\pi i t)\ {\rm d}t =\hat f(n)\ \hat g(n)\ .\cr }

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    @mathusiast: The functions $f$ and $g$ have been assumed continuous by the OP. In this case there is no problem with Fubini. When $f$ and $g$ are only in $L^1$ some work is needed.2013-06-12
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By $f(n)$ etc. I guess you mean the Fourier coefficients, and then it is "yes" depending on which kind of Fourier coefficients we look at.

Suppose $t$ is a 1-periodic continuous function, and consider $\langle f,t \rangle =\int_0^1 f(x)t(x)dx$ then $\langle f*g,t \rangle = \int_0^1 \int_0^1 f(y)g(x-y)dy t(x)dx = \int_0^1f(y)\int_0^1 g(u) t(u+y)du\,dy, $ where we changed the order of integration and used the substitution $u=x-y$. The thing is that if $t$ is a so called character (a continuous group homomorphism from the additive group $\mathbb{R}$ to the circle group) then $t(x+y)= t(x)t(y)$ and then we get $\langle f*g,t \rangle = \int_0^1f(y)\int_0^1 g(u) t(u)t(y)du\,dy =\langle f,t \rangle\cdot \langle g,t \rangle.$ The characters here are $t(x)=e^{2i\pi nx}$ for $n\in\mathbb{Z}$.

If we instead looks at sine and cosine series we have $a_n(f)=\int f(x)\cos 2\pi nx dx$ and $b_n(f)=\int f(x)\sin 2\pi nx dx$ and then we only get relations $a_n(f*g)=\int_0^1f(x) \int_0^1 g(y) \cos 2\pi n (x+y)dydx=\ldots=a_n(f)a_n(g)-b_n(f)b_n(g),$ due to the addition formula for $\cos$ (a simliar expression holds for $b_n$).

Moreover, this is true on any locally compact abelian group.