Is convolution theorem on $l^2(\mathbb{Z})$ valid?

Question

I have a doubt about Fourier transform $F:L^2([0,2\pi])\to l^2(\mathbb{Z})$. If $f,g\in l^2(\mathbb{Z})$ then $f*g\in l^2(\mathbb{Z})$, then, $\mathcal{F}^{-1}(f*g)\in L^2([0,2\pi])$.

Question $\mathcal{F}^{-1}(f*g)=\mathcal{F}^{-1}(f)\mathcal{F}^{-1}(g)$?

Thanks.

Answer 1

I abbreviate $G=[0, 2\pi]$. Note that if $f, g\in L^2(G)$, then $fg\in L^1(G)$ by the Hölder inequality so that $$ (fg)^\wedge(m) = \frac{1}{2\pi}\int_0^{2\pi} f(t)g(t)e^{-imt} dt = \sum_{n\in\mathbb Z} \mathcal F f(n) \overline{\mathcal F({\overline{g} e^{imt}}})(n) $$ using Parseval's identity (here $e^{imt}$ should be understood as the function $t\mapsto e^{imt}$ and applying $\mathcal F$ to the function $t\mapsto g(t)e^{imt}$ in the second equation).

But since $G$ is compact it has finite measure, so that $L^2(G)\subseteq L^1(G)$ and hence $\mathcal F f(n) = \hat f(n)$ and $$ \overline{\mathcal F({\overline{g} e^{imt}}})(n) = \overline{\widehat{\overline{g}e^{imt}}}(n) = \frac{1}{2\pi}\int_0^{2\pi} e^{int} e^{-imt} g(t) dt = \widehat{g}(m-n). $$ Thus combining the two previous calculations we see $$ \widehat{fg}(m) = \sum_{n\in\mathbb Z} \hat f(n) \hat g(m-n) = (\hat f *\hat g)(m). $$

Thus we get $\mathcal F(fg) = (fg)^\wedge = \hat f * \hat g = \mathcal F(f) * \mathcal F(g)$ and thus for $h, k \in\ell^2(\mathbb Z)$ applying this to $f=\mathcal F^{-1} h, g=\mathcal F^{-1} k$ we obtain $$ \mathcal F(\mathcal F^{-1}(h)\mathcal F^{-1}(k)) = h*k $$ so that by applying $\mathcal F^{-1}$ to both sides again we obtain your statement.

Note: This proof generalizes to $G$ being a locally compact abelian group with Plancherel transform $\mathcal F\colon L^2(G)\to L^2(\widehat G)$ but then we have to include an extra step, as in general for non-compact $G$ the function $g$ need not to be in $L^1(G)$, so we can't obtain the second equation. In order to fix this, we may assume $g\in C_c(G)$ and then argue by density of $C_c(G)\subseteq L^2(G)$ among realizing that the equation $\mathcal F(fg) = \mathcal F(f)*\mathcal F(g)$ is continuous in $g$ with respect to the $L^2$-norm (using Hölder, Young, ...).