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A family of functions $k_n(x):[-\pi,\pi]\to \mathbb R$ for $n\in \mathbb N$ is said to be a good kernel if all the following are satisfied:

  1. $\frac{1}{2\pi }\int_{-\pi}^\pi k_n(x) \, \mathrm d x=1$,

  2. $ \int_{-\pi}^\pi |k_n(x)| \, \mathrm d x=O(1)$ where the implied constant is independent of $n$,

  3. for each $\varepsilon>0$ we have $ \lim_{n\to \infty} \int_{\varepsilon<|x|\leq \pi} |k_n(x)| \, \mathrm d x=0$.

This is a standard definition in Fourier analysis, see, for example, the book of Stein--Shakarchi "Fourier analysis" page 48.

For example, if $$ k_n(x) = \sum_{k\in \mathbb Z \\ |k| \leq n} \exp(-\pi k^2/n^2) \mathrm e^{2\pi i k x}$$ then the integral in the third property has size roughly $\exp\left(-\pi \epsilon^2 n^2\right).$

QUESTION: is there a good choice for $k_n$ that makes the limit in $(3)$ be $$ o_{n\to +\infty }\left( \exp\left(-\pi \epsilon^2 n^2\right)\right)?$$

More importantly, is there any literature on this type of question?

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    $\begingroup$ What do you mean by "super fast"? $\endgroup$ Commented Apr 16, 2023 at 18:40
  • $\begingroup$ As a function of $n$ and $\epsilon$. $\endgroup$
    – Dr. Pi
    Commented Apr 16, 2023 at 19:13
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    $\begingroup$ What do you mean here? Would every function whatsoever of $n$ and $\epsilon$ be considered super fast by you? $\endgroup$ Commented Apr 16, 2023 at 19:25
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    $\begingroup$ The problem is, if you take a subsequence $k_{n_i}$ then it will also be a good kernel, and this way you can accelerate the convergence however you want by a diagonal process. So it is not clear what exactly are you asking for? $\endgroup$ Commented Apr 16, 2023 at 19:25
  • $\begingroup$ Thank you Iosif Pinelis and Aleksei Kulikov for the questions! I added some detail and an example to explain a little bit better what I meant. $\endgroup$
    – Dr. Pi
    Commented Apr 16, 2023 at 20:24

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