
> 
> Let $f\in C^1(\mathbb T)=C^1(\mathbb R/\mathbb Z)$ be a function such that
> $$\hat f(k):=\int\_{\mathbb T}f(x)e^{-2\pi ikx}\,dx=0,\qquad \forall k\in\{-N+1,\cdots,-1,0,1,\cdots, N-1\}.$$
> Do we have $\|f\|\_{L^\infty}\leq \frac CN \|f'\|\_{L^1}$ for some $C>0$ indpendent of $f$ and $N$?
> 
> 
> 


We have $\|f\|\_{L^\infty}\leq \frac1{4N} \|f'\|\_{L^\infty}$, see [this MSE post](https://math.stackexchange.com/questions/4389189/if-the-fourier-coefficient-hatfk-of-f-in-c1-mathbb-t-is-zero-for-all/4395619#4395619) for a clever proof. I just wonder, if we change $\|f'\|\_{L^\infty}$ to $\|f'\|\_{L^1}$, do we have the same inequality?


Maybe the decay $\frac1N$ is too fast to be true. If this is not ture, can we find a $C\_N$ such that
$$\|f\|\_{L^\infty}\leq C\_N \|f'\|\_{L^1} \qquad \text{and }\ \ \ \ \lim\_{N\to\infty}C\_N=0?$$


Note that by the fundamental theorem of calculus we obtain $\|f\|\_{L^\infty}\leq \|f'\|\_{L^1}$. Also, 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}.
\end{align\*}


Any help would be appreciated!



If the Fourier coefficient $\hat{f}(k)$ of $f\in C^1(\mathbb T)$ is zero for all $|k|<N$, then $\|f\|_{L^\infty}\leq \frac CN \|f'\|_{L^1}$?
There is no chance. If you could do it, the linear functional $g\mapsto (\int g)(0)$ would have small norm on the corresponding subspace of $L^1$ and, thus, extend to a functional of small norm in the whole $L^1$. Thus, we would have a small in $L^\infty$ function $g$ whose Fourier coefficients are $\frac 1n$ for $|n|>N$, but it would differ only by a trigonometric polynomial from the full series $G(t)=2i\sum\_{n\ge 1}\frac {\sin nt}n$, which has a jump discontinuity at $0$. So $g$ cannot eliminate it, and "Whoosh, fly all our hopes away"...


