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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 $1 \leq p\leq \infty$?

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  • $\begingroup$ Can C depend on p? $\endgroup$ – Yemon Choi Feb 15 '16 at 13:30
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    $\begingroup$ 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 '16 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 '16 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). $\endgroup$ – Christian Remling Feb 23 '16 at 4:30
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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.

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  • $\begingroup$ [deleted comment, I didn't see your comment to the main question] $\endgroup$ – Yemon Choi Feb 22 '16 at 18:53

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