$f,g$ are two continuous functions with period$=1$, are the Fourier coefficients $f*g=f(n)g(n)$? 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!
 A: 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
$$\eqalign{\hat h(n)&=\int_T h(x)\exp(-2n\pi i x)\ {\rm d}x =
\int_T\left(\int_T f(x-t)g(t)\ {\rm d}t\right)\ \exp(-2n\pi i x)\ {\rm d}x \cr
&=\int_{T\times T} f(x-t) g(t)\exp\bigl(-2n\pi i x \bigr)\ {\rm d}x'\ {\rm d}t \cr &=\int_{T\times T} f(x') g(t)\exp\bigl(-2n\pi i(x'+t)\bigr)\ {\rm d}x'\ {\rm d}t \cr &=\int_T f(x')\exp(-2n\pi i x')\ {\rm d}x' \cdot \int_T g(t)\exp(-2n\pi i t)\ {\rm d}t =\hat f(n)\ \hat g(n)\ .\cr
}$$
A: 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. 
