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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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$f*g$ is the convolution of $f$ and $g$, right? –  Paul Jan 6 '12 at 12:09
    
And what are the arguments of $f$ and $g$? It would appear that $f$ and $g$ are continuous (periodic) functions from $\mathbb R$ to $\mathbb R$, and so the standard interpretation of $f(n)$ is the value of $f$ at integer $n$. Since $f$ is periodic with period $1$, $f(n)$ has the same value for all choices of $n$ and similarly for $g(n)$. Can't you think of any other notation for the $n$-th Fourier coefficient of $f$? –  Dilip Sarwate Jan 6 '12 at 12:33
    
This is the convolution of f,g. Do you have a better notion? I'm not familiar with latex –  tomerg Jan 6 '12 at 13:00
    
What is your domain? Since you mentioned period 1, do you mean $f$ and $g$ are functions defined over $\mathbb{R}$? In which case how do you define the convolution? (The "usual" integral doesn't converge at all!) Perhaps you just want $f$ and $g$ to be functions defined on the circle? –  Willie Wong Jan 6 '12 at 13:00
    
f,g:R-->C (must use letters) –  tomerg Jan 6 '12 at 13:05
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2 Answers

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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Could you please comment on why switching the order of integration is allowed here? –  Leo Schmidt Jun 12 '13 at 1:18
    
@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. –  Christian Blatter Jun 12 '13 at 8:03
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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.

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