"Reversed" Bernstein Inequality I'm studying harmonic analysis by myself, and I read some online notes that introduce the Bernstein inequality. One of them mention a reversed form of the Bernstein inequality, which is stated below:
Let $\mathbb{T} = \mathbb{R} / \mathbb{Z} = [0,1]$ be the one-dimensional torus. Assume that a function $f \in L^{1}(\mathbb{T})$ satisfies $\hat{f}(j) = 0$ for all $|j| < n$ (vanishing Fourier coefficients). Then for all $1 \leq p \leq \infty$, there exists some constant $C$ independent of $n,p$ and $f$, such that
$$\|f'\|_{p} \geq Cn\|f\|_{p}$$
It seems that an easier problem can be obtained by replacing $f'$ with $f''$ in the above inequality. The easier problem is addressed in the MO post below:
Does there exist some $C$ independent of $n$ and $f$ such that $ \|f''\|_p \geq Cn^2 \| f \|_p$, where $1 \leq p\leq \infty$?
However, it seems that the trick of convex Fourier coefficients used in the post above no longer applies to the harder problem (lower bounding the norm of the first derivative). Any suggestions/ideas?
 A: Let me define the Fourier multiplier $\Omega_N (D)$ by
$$
(\Omega_N(D) u)(x)=\text{Fourier}^{-1}\bigl(\hat u(\xi)\mathbf 1 (\vert \xi\vert \ge N)\bigr).
$$
We define then the operator  $A_N(D)=D\Omega_N(D)$
(also a Fourier multiplier). We have obviously
$$
\Vert A_N(D) u\Vert_{L^2}\ge N \Vert \Omega_N(D) u\Vert_{L^2},\quad\text{i.e.}\quad
\Vert D u_N\Vert_{L^2}\ge N \Vert u_N\Vert_{L^2}\quad\text{ with $u_N=\Omega_N(D) u$.}
$$
Let $p$ be in $(1,+\infty)$. We have
$$
u_N= \underbrace{\vert D\vert^{-1} N\Omega_N(D)}_{
\substack{\text{bounded operator on } L^p}}N^{-1}\vert D\vert u_N,
$$
so that
$$
\Vert u_N\Vert_{L^p}\le C_p\Vert N^{-1}\vert D\vert u_N\Vert_{L^p}
\le N^{-1}\tilde C_p\Vert D u_N\Vert_{L^p},
$$
where the last inequality is due to the $L^p$ continuity of the Hilbert transform (the Fourier multiplier $\text{sign} D$) since
$
\vert D\vert=D \text{ sign} D
$.
Above we have used that
$\vert D\vert^{-1} N\Omega_N(D)$ is the Fourier multiplication by
$$
\vert \xi\vert^{-1} N\mathbf 1 (\vert \xi\vert \ge N),
$$
and the Hörmander-Mihlin multiplier theorem gives $L^p$ boundedness.
This does not work for $p=1,\infty$.
