Fourier Coefficients of the Product of Two Functions – Fourier Analysis

fourier analysisfourier seriesinner-products

Given two functions $f,g\in L^2(\mathbb{T})$, I have to prove that the Fourier coefficients of $fg$ are given by

$$\hat{fg}(n)=\sum_{k\in{Z}}\hat{f}(n-k)\hat{g}(k)$$

and that this series converges uniformly.

Thank you in advance for any help.

Best Answer

By Cauchy-Schwarz and Parseval: $$ \sum_{k\in\mathbb{Z}}|\hat{f}(n-k)||\hat{g}(k)|\leq (\sum_{k\in\mathbb{Z}}|\hat{f}(n-k)|^2)^{\frac{1}{2}}(\sum_{k\in\mathbb{Z}}|\hat{g}(k)|^2)^{\frac{1}{2}}=\|f\|_2\|g\|_2 $$ so the rhs converges absolutely. By Cauchy-Schwarz again, but in $L^2$, the pointwise product $fg$ is in $L^1$, so the lhs is well-defined.

Note that these are both bilinear forms on $L^2\times L^2$ bounded by $\|f\|_2\|g\|_2$. To prove that they coincide, it suffices to do it for a dense set. By Parseval, the trigonometric polynomials form such a set.

To verify this for the trigometric polynomials, it suffices to do it for $p(t)=e^{irt}$ and $q(t)=e^{ist}$ for any $r,s\in\mathbb{Z}$. Now $$ (pq)(t)=e^{i(r+s)t}\qquad\Rightarrow\qquad \widehat{pq}(n)=\delta(n,r+s) $$ and $$ \sum_{k\in\mathbb{Z}}\hat{p}(n-k)\hat{q}(k)=\sum_{k\in \mathbb{Z}}\delta(n-k,r)\delta(k,s)=\delta(n-s,r)=\delta(n,r+s). $$

Note: I realize I did it for $L^2([0,2\pi])$. To adapt this to $L^2(\mathbb{T})$, just replace $p$ and $q$ by $z^r$ and $z^s$.

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