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Let $n$ be a odd number. Does there exist any $2\pi$-periodic continuous function $f :\mathbb{R}\to \mathbb{R}$ such that the following points simultaneously hold?

1- $-n\lneqq f_{\min}$ (where $f_{\min}$ is the minimum of $f$).
2- If $|k|\leq \frac{n-1}{2}$, the Fourier coefficients $\hat{f}(k)=\sec\frac{k\pi}{n}$.

We recall that the Fourier series of $f$ is just $\sum_{k\in \mathbb{Z}} \hat{f}(k)e^{ikx}$ with Fourier coefficients $\hat{f}(k)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-ikt}dt$.

p.s. However I am concerned the above question, the following general form can also be considered. For a given a sequence of real numbers $\{a_k\}_{k=-n}^n$ and $\gamma\in \mathbb{R}$, does there exist any $2\pi$-periodic continuous function $f :\mathbb{R}\to \mathbb{R}$ such that the following points simultaneously hold?

1- $\gamma \leq f_{\min}$ (where $f_{\min}$ is the minimum of $f$).
2- If $|k|\leq n$, the Fourier coefficients $\hat{f}(k)=a_k$.

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  • $\begingroup$ Do you mean $|k|\le \frac{n-1}2 $? $\endgroup$ Commented Jan 6, 2023 at 9:28
  • $\begingroup$ Yes, thanks and corrected. $\endgroup$
    – ABB
    Commented Jan 6, 2023 at 9:29

1 Answer 1

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I define the Fourier series as $$f(x) = \tfrac{1}{2}c_0 + \sum_{k=1}^{\infty} c_k \cos kx.$$ The question is a bit unclear on whether a functional form is needed for all $n$, but for a limited range $1\leq n\leq 27$ this works $$f(x)=\tfrac{1}{2}+\sum_{k=1}^{(n-1)/2}\frac{\cos (kx)}{\cos(k\pi/n)}.$$

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  • $\begingroup$ Thanks a lot. If we write, $2\cos(kx)=e^{ikx}+e^{-ikx}$, then it seems that $\hat{f}(k)=\hat{f}(-k)=\frac{1}{2}\sec\frac{k\pi}{n}$ and so the minimum of $f$ is changed. $\endgroup$
    – ABB
    Commented Jan 6, 2023 at 11:23
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    $\begingroup$ I use the definition of Fourier coefficients from Wikipedia: $f(x) = \tfrac{1}{2}c_0 + \sum_{k=1}^{\infty} c_k \cos kx$ --- you may want to specify your definition in the post $\endgroup$ Commented Jan 6, 2023 at 11:26
  • $\begingroup$ The example that you mentioned is nice and hopefully works for all $n$ (with probably minor change). $\endgroup$
    – ABB
    Commented Jan 6, 2023 at 11:29
  • $\begingroup$ I am extremely interested in the reason of the off vote! Have you ever read it only once? It is logical to write at least one sentence to give negative point. $\endgroup$
    – ABB
    Commented Jan 6, 2023 at 14:32
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    $\begingroup$ @ABB --- voting is anonymous, for what it's worth I did not down vote. $\endgroup$ Commented Jan 6, 2023 at 14:37

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