TITLE: Does there exist some $C$ independent of $n$ and $f$ such that $ \|f''\|_p \geq Cn^2 \| f \|_p$, where $1 \leq p\leq \infty$?
QUESTION [14 upvotes]: Let $f$ be a trigonometric polynomial on the circle $\mathbb{T}$ with $\hat{f}(j) = 0$ for all $j \in \mathbb{Z}$ with $\lvert j \rvert < n$. Does there exist some $C$ independent of $n$ and $f$ such that
$$
\|f''\|_p \geq Cn^2 \| f \|_p,
$$
where $1 \leq p\leq \infty$?

REPLY [3 votes]: Here is more pedestrian argument. If $T_{n}$ is a trigonometric polynomial of degree at most $n$ with total mass $\frac{1}{2\pi}\int_{-\pi}^{\pi}T_{n}=1$ then $f*T_{n}=0$, and in particular, 
$$
f(x) =\frac{1}{2\pi}\int_{-\pi}^{\pi}(f(x)-f(x-s))T_{n}(s)ds.
$$
Therefore, by the triangle inequality 
$$
\|f\|_{p} \leq \frac{1}{2\pi}\int_{-\pi}^{\pi}\|f(x)-f(x-s)\|_{L^{p}(dx)}|T_{n}(t)|dt \leq \|f'\|_{p}\frac{1}{2\pi}\int_{-\pi}^{\pi}|s||T_{n}(s)|ds,
$$
where the inequality $\|f(x)-f(x-s)\|_{L^{p}(dx)} \leq |s| \|f'\|_{p}$ follows, for instance from Schur test applied to the operator $(Af')(x)=\int f'(t) 1_{[x-s,x]}(t)dt$. 
Now how can we make $\frac{1}{2\pi}\int_{-\pi}^{\pi}|s||T_{n}(s)|ds$ of order $\frac{1}{n}$? Let us be not too demanding and seek for $T_{n}$ among even nonnegative  trigonometric polynomials to reduce the matters to $\int_{0}^{\pi}sT_{n}(s)ds$. 
One immediate choice is Fejer kernel 
$$
k_{n}(s) = \frac{1}{n}\left(\frac{\sin(\frac{ns}{2})}{\sin(\frac{s}{2})}\right)^{2}.
$$
Now $k_{n}(s) \asymp  n$ on $[0,\frac{1}{n}]$, and $k_{n}(s)<C \frac{1}{n}\frac{1}{s^{2}}$ on $[\frac{1}{n}, \pi]$, therefore $\int_{0}^{\pi}sk_{n}(s)ds \leq C' \frac{\log(n)}{n}$. Well, not too bad but not exactly what was requested. 
What else can we do? Let us look at $k_{n}^{2}(s)$. It is even nonnegative trigonometric polynomial of degree $2n$. A small jump in degree is okay (we can just start from $f$ of degree $\geq 2n$). Since $k_{n}^{2}(s) \asymp  n^{2}$ on $[0, \frac{1}{n}]$ its total mass is at least $\geq n$. Then 
$$
\frac{k_{n}^{2}}{n} \asymp n \quad \text{on}\quad [0, \frac{1}{n}], \quad \text{and} \quad \frac{k_{n}^{2}}{n} \leq C\frac{1}{n^{3}} \frac{1}{s^{4}} \quad \text{on} \quad [\frac{1}{n}, \pi]
$$
therefore $\int_{0}^{\pi} s \frac{k^{2}_{n}}{n}\leq C' \frac{1}{n}$ voila!