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 $\|f\|_{L^\infty}\leq \frac CN \|f'\|_{L^\infty}$ for some $C>0$ indpendent of $f$ and $N$.
The best constant $C$ is $\frac14$, as shown by this post, in which the proof uses a very "hard" theorem due to Vaaler.
I'm trying to prove the decay rate $N^{-1}$ and I don't care about the best constant $C$. I'm looking for an easier proof.
First Try. By Cauchy inequality and the Plancherel identity, \begin{align*} \|f\|_{L^\infty}&\leq\sum_{|k|\geq N}|\hat f(k)|=\sum_{|k|\geq N}\frac{\left|\widehat{f'}(k)\right|}{2\pi |k|}\\ &\leq \frac1{2\pi}\left(\sum_{|k|\geq N}\left|\widehat{f'}(k)\right|^2\right)^{1/2}\left(\sum_{|k|\geq N}\frac1{|k|^2}\right)^{1/2}\\ &\leq \frac1{2\pi}\left\|f'\right\|_{L^2}\left(\frac2N\right)^{1/2}=\frac C{\sqrt N}\left\|f'\right\|_{L^2}\\ &\leq \frac C{\sqrt N}\left\|f'\right\|_{L^\infty}\ \ . \end{align*}
The deacy rate on $N$ is not good enough to get the desired result.
Second Try. (Also failed.) As Sarvesh Ravichandran Iyer suggested in the comments, using some ideas in the proof that I quoted above, I've made some progress. In this post, let $\psi(x)=x-1/2, x \in [0,1]$, then $$\psi(x)=-\frac{1}{\pi}\sum_{k \ge 1}\frac{\sin 2\pi kx}{k}, 0<x<1$$ and $\psi*f'=-f$. 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 $||\eta||_1 \le \frac{C}{N}$ then we are done. A natural candidate is $$\eta(x)=\psi(x)+\frac{1}{\pi}\sum_{1\leq k\leq N-1}\frac{\sin 2\pi kx}{k}=-\frac{1}{\pi}\sum_{k \ge N}\frac{\sin 2\pi kx}{k}.$$ Now, the problem reduces to show that $||\eta||_1 \le \frac{C}{N}$ for this special function $\eta$, where $C>0$ is independent of $N$. To explore the oscillation in this summation, we can use the partial sum method. Let $S_M(x)=\sum_{k=N}^M \sin(2\pi kx)$, then $$S_M(x)=-\frac{\cos((2M+1)\pi x)-\cos((2N-1)\pi x)}{2\sin \pi x}=\frac{\sin((M+N)\pi x)\sin((M-N+1)\pi x)}{\sin\pi x}.$$ The partial sum method applied on $\eta$ gives that $$\eta(x)=-\frac1\pi\sum_{k=N}^\infty \frac1{k(k+1)}S_k(x).$$ Note that near $x=0$, the denominator is small, hence, naturally we split the integral into two parts: $\int_0^{1/2}|\eta|=\int_0^\delta+\int_\delta^{1/2}$. By Plancherel we have $$\int_0^\delta |\eta(x)|\,dx\leq \delta^{1/2}\left(\int_0^\delta|\eta(x)|^2\,dx\right)^{1/2}\leq C\delta^{1/2}\left(\sum_{k=N}^\infty\frac1{k^2}\right)^{1/2}\leq C\sqrt{\frac\delta N}.$$ To get the correct decay rate $N^{-1}$, it seems that $\delta=1/N$ is a good choice. It remains to estimate $\int_\delta^{1/2}|\eta|$, we have $$\int_\delta^{1/2}|\eta(x)|\,dx\leq C\sum_{k=N}^\infty \frac1{k(k+1)} \int_\delta^{1/2}\frac1{|\sin(\pi x)|}\,dx\leq C\frac {|\log \delta|}{N}=C\frac{\log N}N.$$ Therefore, $\|\eta\|_1\le C\frac{\log N}N$. This implies that $\|f\|_{L^\infty}\leq C\frac{\log N}N \|f'\|_{L^\infty}$, which is better than my first try, but not satisfactory either.
Any help would be appreciated!