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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?2012-01-06
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    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 $f$?2012-01-06
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    This is the convolution of f,g. Do you have a better notion? I'm not familiar with latex2012-01-06
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    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?2012-01-06
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    f,g:R-->C (must use letters)2012-01-06

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