Properties of $L^1([0,1]) \to \ell^\infty(\mathbb Z)$

Question

I want to learn more about the map $$\varphi: L^1([0,1]) \to \ell^\infty(\mathbb Z), \quad \varphi(f)_n = \int_0^1 f(x) \exp(-2\pi nix) dx. $$ It is linear and continuous since we have $||\varphi(f)||_{\infty} \le ||f||_1$. Here I learnt that it is injective. Is it an isometry? What does the cokernel look like?

Answer 1

It is not an isometry. Consider, for example, $$ f(x) = \sum_{n=-N}^N \exp(2\pi i n x) = \frac{\sin((2N+1)\pi x)}{\sin(\pi x)} ,$$ where the right hand side comes from summing the geometric series. Then it can be shown that ${\|f\|}_1$ is of the order of $\log N$, but clearly ${\| \varphi (f)\|}_\infty = 1$.

By the Riemann-Lebesgue Lemma https://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma, we can see that $\varphi(f) \in c_0(\mathbb Z)$, and it is easily seen that the range of $\varphi$ is dense in $c_0(\mathbb Z)$.