$\sup_{k\in\mathbb{Z}} k \cdot \int_0^{2\pi} f(t)e^{ikt} \, dt <\infty$

Question

Given a measurable function $f$ on $(0,2\pi)$. If $f$ is bounded and square integrable, then can we say that $$\sup_{k\in\mathbb{Z}} k \cdot \left|\int_0^{2\pi} f(t)e^{ikt} \, dt \right| <\infty ? $$

Answer 1

What you are looking at is the sequence $k |\langle f, e^{ikt} \rangle|$, where $\langle f, g \rangle = \int_0^{2\pi} fg$ is the usual inner product on $L^2([0, 2\pi])$. This space is famously isometrically isomorphic to the sequence space $\ell^2(\mathbb{Z})$, where $f \in L^2([0, 2\pi])$ maps to $(\langle f, e^{ikt} \rangle)_{k \in \mathbb{Z}}$ is an example of such an isometric isomorphism.

So, the question is, is it true that $\sup_{k \in \mathbb{Z}} k |x_k| < \infty$ for square-summable doubly-infinite sequences $(x_k)$? The answer is "no". For example, take $x_k =|k|^{-\frac{3}{4}}$. Or, taking it back to functions, a counter-example would be $$f(t) = \sum_{k \in \mathbb{Z}} |k|^{-\frac{3}{4}} e^{ikt}.$$