If the Fourier coefficient $\hat{f}(k)$ of $f\in C^1(\mathbb T)$ is zero for all $|k|
Let $f\in C^1(\mathbb T)=C^1(\mathbb R/\mathbb Z)$ be a function such that
$$\hat f(k):=\int_{\mathbb T}f(x)e^{-2\pi ikx}\,dx=0,\qquad \forall k\in\{-N+1,\cdots,-1,0,1,\cdots, N-1\}.$$
Show that $\|f'\|_{L^\infty}\geq 4N\|f\|_{L^\infty}$.
Here is my attempt. Since for $f\in C(\mathbb T)$, the Cesàro mean
$$\sigma_n f(x):=\frac{1}{n+1}\sum_{j=0}^{n}S_j f(x):=\frac{1}{n+1}\sum_{j=0}^{k}\sum_{k=-j}^j\hat f(k)e^{2\pi ikx}$$
converges to $f$ in $L^\infty$, it suffices to estimate on $\|\sigma_n f\|_{L^\infty}$ and then pass the limit. Integration by parts gives $\widehat{f'}(k)=2\pi ik\hat f(k)$, and so for any $n\geq N$ we have
$$\sigma_nf(x)=\frac{1}{n+1}\sum_{j=0}^{n}S_j f(x)=\frac{1}{n+1}\sum_{j=N}^{n}\left(\sum_{k=-j}^{-N}+\sum_{k=N}^j\right)\frac{\widehat{f'}(k)}{2\pi ik}e^{2\pi ikx}.$$ Now I don’t know how to continue.
Any hints are welcome!
 A: We prove that if $f$ periodic and differentiable on $[0,1]$ with $f'$ bounded (eg $f'$ continuous) and $f(x)=\sum_{|k| \ge N}a_ke^{2\pi ikx}$ is the Fourier series of $f$ where $N \ge 1$ is a fixed integer, then $||f'||_{\infty} \ge 4N||f||_{\infty}$.
Let $\widehat{f'}(k)=b_k=2\pi ika_k$ so $a_k=\frac{b_k}{2\pi i k}, |k| \ge N$
We use the well known Fourier series for $\psi(x)=x-1/2, x \in [0,1]$ and a result of Vaaler regarding its $L^1$ approximation by trigonometric polynomials of degree at most $N-1$
It is well known and not hard to prove that $\psi(x)=-\frac{1}{\pi}\sum_{k \ge 1}\frac{\sin 2\pi kx}{k}, 0<x<1$ (with bounded convergence). But since $\sin 2\pi kx=\frac{1}{2i}e^{2\pi ikx}-\frac{1}{2i}e^{-2\pi ikx}$ so $\psi(x)=-\frac{1}{2\pi i}\sum_{|k|\ne 0}\frac{e^{2\pi ikx}}{k}, 0<x<1$, one has that the Fourier coefficients of $\psi*f'$ are precisely $-\frac{b_k}{2\pi ik}, |k| \ge N$ hence $\psi*f'=-f$ ae by general theory (integrable functions with same Fourier coefficient are equal ae)
But now if we perturb $\psi$ by any trigonometric polynomial of degree at most $N-1$ and we get an $\eta=\psi-P_{N-1}$ we still have $\eta*f'=-f$ hence $||f||_{\infty}=||\eta*f'||_{\infty} \le ||\eta||_1||f'||_{\infty}$ and in particular if we find $||\eta||_1 \le \frac{1}{4N}$ we are done
But now Vaaler theorem (Theorem 17, p 208, paper "Some extremal functions in Fourier analysis" linked here) gives precisely this result by exhibiting a trigonometric polynomial $\psi*i_{N-1}$ of degree $N-1$ st $\eta_{N-1}=\psi-\psi*i_{N-1}$ satisfies $||\eta_{N-1}||_1 = \frac{1}{4N}$ (and moreover showing that this the best we can do so for any other trigonometric polynomial of degree at most $N-1$ the integral above is at least $\frac{1}{4N}$ with equality only for $\psi*i_{N-1}$), so the problem is solved!
