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Given an $L^2(\Bbb R)$ function $f$, let $c_n(f)$ be its $n$th Fourier coefficient. Define an isometry $T: L^2 \to \ell^2$ by $Tf = (c_0(f),c_1(f),\ldots)$. Let $F$ be the Fourier transform operator; it's an isometry on $L^2$. So how does $T\circ F$ act on $\ell^2$? What is it doing to the Fourier coefficients?

Here, $F(f)(s) = \int_{-\infty}^\infty e^{-2\pi i st}f(t) dt$.

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  • $\begingroup$ Domain is $\Bbb R$, the formula is $F(f)(s) = \int_{-\infty}^\infty e^{-2\pi i st}f(t) dt$. $\endgroup$
    – user541020
    Commented Nov 4, 2019 at 23:25
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    $\begingroup$ To clarify, are you asking for whether there is an operator $G$ such that $G \circ T = T \circ F$, and if so to describe what $G$ is? (confusion coming from the fact $T \circ F$ doesn't act on $\ell^2$ formally) $\endgroup$
    – Brian Moehring
    Commented Nov 4, 2019 at 23:34
  • $\begingroup$ Can you give a formula for $c_k(f)$ as well. When I went to school elements of $L^2(\mathbb T)$ had Fourier coefficients in $\ell^2$, not elements of $L^2(\mathbb R)$. What am I missing? $\endgroup$
    – kimchi lover
    Commented Nov 5, 2019 at 12:51

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