Fourier coefficients assume a maximum and minimum?

Question

Let $f:\mathbb R\to\mathbb R$ be continuously differentiable and periodic with period $2\pi$.

The Fourier coefficients are defined by $$\hat f_n=\int_{-\pi}^\pi f(x)\exp(-inx)dx$$

My questions: Is $\widehat {f}_n$ bounded? Does $\widehat {f}_n$ assume a maximum or minimum?

My attempt: Since $f'$ is continuous on the compact interval $[-\pi,\pi]$, $f'$ assumes a maximum and minimum, so we get $|f'(x)|\leq C$. So we get $$\left|\widehat {f}_n\right|=\left|\frac{1}{2\pi in}\int_{-\pi}^\pi f'(x)\exp(-inx)\,dx\right|\leq\frac{1}{2\pi |n|}\int_{-\pi}^\pi C \, dx=\frac C{|n|} $$ So $\widehat f_n$ is bounded.

But does the Fourier coefficient of a continuous function assume a maximum/minimum or do we get just a supremum/infimum? I couldn't have thought about a counterexample so I would say yes but I am really unsure about it.

Anybody has an idea how to show it or a counterexample?

Answer 1

Observe that $\hat f_n$ may be a complex number, and that $\mathbb{C}$ is not an ordered field. In particular, the supremum and infimum of $\{\hat f_n\}$ are not defined in general.

You have shown that $\{\hat f_n\}$ is bounded and $\lim_{|n|\to\infty}\hat f_n=0$. Fron this you should be able to prove that $\{|\hat f_n|\}$ attains its maximum, but unless some coefficient equals cero, not its infimum.

Answer 2

Indeed, this will be the case for an arbitrary integrable function $f$. The Riemann-Lebesgue Lemma implies that $\lim_{n \to \pm \infty} \hat f_n = 0$, so that $\left\{\left|\hat f_n\right|\right\}_{n \in \mathbb{\mathbb{Z}}}$ must be a bounded set.