Let $f$ be a trigonometric polynomial on the circle $\mathbb{T}$ with $\hat{f}(j) = 0$ for all $j \in \mathbb{Z}$ with $\lvert j \rvert < n$. I want to show $$ \|f''\|_p \geq Cn^2 \| f \|_p. $$ for some $C$ independent of $n$ and $f$ and $1 \leq p\leq \infty$.
I also have a result that may be useful: if $(a_n)_{n \in \mathbb{Z}}$ is an even ($a_n = a_{-n}$) sequence of nonnegative numbers with $a_n \to 0$ and $$ a_{n+1} + a_{n-1} - 2 a_n \geq 0 \quad \forall n > 0, $$ then there is an $f \in L^1(\mathbb{T})$ with $f \geq 0$ and $\hat{f}(n) = a_n$.
I am struggling to find an entrance into the problem. I would most appreciate a hint that allows me to start. I have tried finding a convolution kernel that would give information via Young's inequality, but this seems unlikely. I would appreciate references that contain more detail as well.
Update
I have the weaker result $\| f'' \|_p \geq Cn^{3/2} \| f \|_p$.
My method uses the observation $f = g \ast f''$ where $\hat{g}(j) = 1/j^2$ and Young's inequality. I can take $\hat{g}(j) = 0$ for $\lvert j \rvert < n$, and bounding $\| g \|_1$ by $\|g \|_2 = \| \hat{g} \|_{\ell^2}$ gives my weaker result.
A correct method must then bound $\| g \|_1$ directly, perhaps using some sort of positivity coming from the stated result on convex Fourier coefficients. I am working on this now, though I may bounty soon.
A generalization of this result: $\| f' \|_p \geq Cn \|f \|_p$ holds, but this is harder to prove. This exercise comes from the book by Schlag and Muscalu.
If $(a_n)_{n \in \mathbb{Z}}$ is an even sequence of nonnegative numbers with $$ a_{n+1} + a_{n-1} - 2a_n \geq 0 \quad \forall n > 0, $$ then there exists $g \in L^1(\mathbb{T})$ with $g \geq 0$ and $\hat{g}(n) = a_n$. This is lemma 1.12 in Classical and Multilinear Harmonic Analysis Vol 1 by C. Muscalu and W. Schlag. The desired function is $$ g = \sum_{n=1}^\infty n (a_{n+1} + a_{n-1} - 2a_n) K_n $$ where $K_n$ is the Fejér kernel.
Define the sequences $(a_{n,j})_{j=0}^\infty$ by $$ a_{n,j} = \begin{cases} \frac{1}{n^2} + \frac{2(n-j)}{n^3},& \text{if } j < n\\ \frac{1}{j^2}, & \text{if } j \geq n \end{cases} $$ for each $n \in \mathbb{N}$. Then (extending to $j \in \mathbb{Z}$ by $a_{n,(-j)} = a_{n,j}$) we can use the lemma to find $g_n \in L^1(\mathbb{T})$ with $g_n \geq 0$ and $\hat{g}_n(j) = a_{n,j}$.
By the monotone convergence theorem, we have $$ \|g_n \|_1 = \sum_{j=1}^\infty j(a_{n,(j+1)} + a_{n,(j-1)} - 2 a_{n,j}). $$ A computation will show that $\| g_n \|_1$ is dominated by $n^{-2}$. Furthermore, for any trigonometric polynomial $f$ with $\hat{f}(j) = 0$ for all $| j | < n$, we have $$ f = g_n \ast f'' $$ so that Young's inequality finishes the proof.
Someone posted this question at this Math Overflow thread, and another proof has been posted there.