# "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?

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 $$p=1,\infty$$.