# How does the Fourier transform act on $\ell^2$?

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

• Domain is $\Bbb R$, the formula is $F(f)(s) = \int_{-\infty}^\infty e^{-2\pi i st}f(t) dt$. Commented Nov 4, 2019 at 23:25
• 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) Commented Nov 4, 2019 at 23:34
• 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? Commented Nov 5, 2019 at 12:51