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 $1 \leq p\leq \infty$?

  • $\begingroup$ Can C depend on p? $\endgroup$
    – Yemon Choi
    Feb 15, 2016 at 13:30
  • 2
    $\begingroup$ The more general inequality $\| f' \|_p \geq Cn \| f \|_p$ should in fact hold. This is problem 1.8 in the first volume of Classical and Multilinear Harmonic Analysis by C. Muscalu and W. Schlag. I have a thread about this on Math StackExchange which I will update accordingly. $\endgroup$
    – Eric Thoma
    Feb 22, 2016 at 6:46
  • $\begingroup$ @ChristianRemling If if helps with justification of the edit, I would be eager to see the extension of your argument to the more general inequality. I have encountered an obstacle in the extension of my argument, as the sequence $a_{n,j}$ is no longer even. $\endgroup$
    – Eric Thoma
    Feb 23, 2016 at 4:15
  • $\begingroup$ @EricThoma: Actually, I've now discovered a problem with my answer and I've deleted (wasted too much time on this already). I'm not finding an easy argument why approximations of $\sum e^{ikx}/k$ should have an $L^1$ error not worse asymptotically than approximations of characteristic functions (though "philosophically" it's clear this has to be right). $\endgroup$ Feb 23, 2016 at 4:30

2 Answers 2


I wish to add another proof based on the following result.

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.

  • $\begingroup$ [deleted comment, I didn't see your comment to the main question] $\endgroup$
    – Yemon Choi
    Feb 22, 2016 at 18:53

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!


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