I am attempting to prove the following: If $f$ is a trigonometric polynomial such that $\hat{f}(j) = 0$ for $|j|<n$ then there exists absolute constant $C$ such that $$Cn\|f\|_p \le \|f'\|_p$$ where $C$ is independent of $p$, $f$ and $n$.
There is the simpler result, saying $$Cn^2\|f\|_p \le \|f''\|_p.$$
Here is my proof for this, and we can see where it fails in the first case. The proof relies on the following lemma:
Let ${a_n}_{n \in \mathbb{Z}}$ be an even sequence of nonnegative numbers that tend to zero, which is convex in the following sense: $a_{n+1} +a_{n−1} −2a_n\ge 0 \forall n>0.$ Then there exists $f \in L^1(\mathbb{T})$ with $f \ge 0$ and $\hat{f}(n) = a_n.$
So for the $n^2$ case: we want a convex, even sequence $a_j \searrow 0$ which will give rise to $h \in L^1$ such that $f = h \ast f''$. We see that $\widehat{h \ast f''} =-4\pi^2j^2 \hat{h}(j)\hat{f}(j)$. So, we take $$a_{n,j} = \begin{cases} \frac{1}{j^2} & |j|\ge n \\ \frac{1}{n^2} & |j| < n.\end{cases}$$ It is clear that the convexity condition is satisfied by a picture. Then let $g_n$ be the function guaranteed by the lemma with $\hat{g}_n(j) = a_{n,j}$ for all $j$. We see $$\|g_n\|_1 = \hat{g}(0)= \frac{1}{n^2}.$$ Then take $h_n=\displaystyle -\frac{1}{4\pi^2}g_n$, so that $\widehat{h \ast f''}(j) = \hat{f}(j)$ for all $j$ (recall that $\hat{f}(j) =0$ for $|j|<n$) and so $f = h \ast f''$. Then Young's inequality $$\|f\|_p = \|h_n \ast f''\|_p \le \|h_n\|_1 \|f''\|_p = \frac{1}{4\pi^2}\|g_n\|_1 \|f''\|_p = \frac{1}{4\pi^2n^2}\|f''\|_p$$ and we are done.
But this won't work for the other case, since the coefficients we will want to define won't be nonnegative. Could anyone please provide a hint about how to proceed?
This problem comes from Schlag's book. Thank you!
EDIT: Ok, my "proof" of the $n^2$ case is actually wrong. When I drew the picture I conflated $1/k^2$ with $k^2$. So, this is not a convex sequence. There is a simple fix, as in the linked post.
As the OP did, we can define the function $K(j)=\frac{1}{j}$ and $K_l(j)=K(j)\varphi_l$, where $l\ge 0$, $\varphi_l(s)=\varphi(2^{-l}s)$ are smooth symmetric functions supported in $|s|\sim 2^l$ and satisfying $\sum_{l\ge 0}\varphi_l=1$. We are interested in the function $K^n = K\sum_{l\ge \log_2n}\varphi_l$. Hence, our desired bound is $\lVert g*(K^{n})^\vee\rVert_p\le \frac{C}{n}\lVert g\rVert_p$, where $K_l^\vee(x) =\sum_j\varphi_l(j)\frac{e^{ijx}}{j}$ for $-\pi\le x\le \pi$.
By the triangle inequality and Young's inequality for convolutions we get $\lVert g*(K^{n})^\vee\rVert_p\le \sum_{l\ge \log_2n}\lVert g*K_l^\vee\rVert_p\le \lVert g\rVert_p\sum_{l\ge 0}\lVert K_l^\vee\rVert_1$, hence we must bound $\lVert K_l^\vee\rVert_1$.
We have that $|K_l^\vee(x)|\le \sum_j\frac{|\varphi_l(j)|}{j}\le C$, for some constant independent of $l$. Note that $e^{i(j+1)x}+e^{i(j-1)x}-2e^{ijx}=e^{ijx}(e^{ix}+e^{-ix}-2)$, hence
$$\begin{align}|(2^lx)^2K_l^\vee(x)|&=\frac{|2^lx|^2}{|e^{ix}+e^{-ix}-2|}\Big|\sum_j \varphi_l(j)\frac{e^{i(j+1)x}+e^{i(j-1)x}-2e^{ijx}}{j}\Big| \\ &\le \frac{2^lx^2}{|e^{ix}+e^{-ix}-2|}\sum_j \left|\frac{\varphi(2^{-l}(j+1))}{2^{-l}(j+1)}+\frac{\varphi(2^{-l}(j-1))}{2^{-l}(j-1)}-2\frac{\varphi(2^{-l}j)}{2^{-l}j}\right|. \end{align}$$
Since $x^2/|e^{ix}+e^{-ix}-2|\le 1$ and
$$\begin{align}\left|\frac{\varphi(2^{-l}(j+1))}{2^{-l}(j+1)}+\frac{\varphi(2^{-l}(j-1))}{2^{-l}(j-1)}-2\frac{\varphi(2^{-l}j)}{2^{-l}j}\right|&\le \sup_s\left|\Big(\frac{\varphi(2^{-l}t)}{2^{-l}t}\Big)''(s)\right| \\ &\le C2^{-2l}, \end{align}$$
we get $|(2^lx)^2K_l^\vee(x)|\le C$ and then
$$ |K_l^\vee(x)|\le C \begin{cases} 1 & \text{for } |x|\le 2^{-l}, \\ \frac{1}{(2^lx)^2} & \text{for } |x|>2^{-l}. \end{cases}$$ We get that $\lVert K_l^\vee\rVert_1\le C2^{-l}$.
To conclude, replace this above to get
$$\lVert g*(K^{n})^\vee\rVert_p\le C\lVert g\rVert_p\sum_{l\ge \log_2n}2^{-l}\le \frac{C}{n}\lVert g\rVert_p,$$
which is what we wanted. I hope there is no mistake.