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Let $\varepsilon>0$. I was interested in understanding the justification of defining the following function $\phi$ via its Fourier transform, satisfying the following properties:

(1) $\widehat{\phi}\in C^\infty(\mathbb{R})$

(2) $\widehat{\phi}(\xi)=1$ for $\xi\in[-\pi+\varepsilon,\pi-\varepsilon]$

(3) supp$(\widehat{\phi})\subset[-\pi-\varepsilon,\pi+\varepsilon]$

(4) $\widehat{\phi}$ goes from 1 to 0 monotonically on the intervals $[-\pi-\varepsilon,-\pi+\varepsilon]$ and $[\pi-\varepsilon,\pi+\varepsilon]$.

(5) $\underset{n\in\mathbb{Z}}\sum\widehat{\phi}(\xi-2\pi n)=1$ for every $\xi\in\mathbb{R}$.

The Fourier Transform convention used was $$\widehat{f}(\xi)=\int_\mathbb{R}f(x)e^{-ix\xi}dx.$$ I am sure there is no problem with it, but I don't quite understand how the construction goes. I would appreciate any explanation. Thank you very much!

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  • $\begingroup$ Do you understand how to do the construction if you do not have to satisfy property 5 necessarily? $\endgroup$ – Ian Dec 13 '15 at 22:18
  • $\begingroup$ No, I am not quite sure how that construction works either... $\endgroup$ – Johnny T. Dec 14 '15 at 1:45
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    $\begingroup$ Look up bump functions and approximate identities. Basically, the function that you want (property 5 aside) is the convolution of an indicator function with an appropriate bump function. $\endgroup$ – Ian Dec 14 '15 at 2:01
  • $\begingroup$ Related $\endgroup$ – Giuseppe Negro Dec 16 '15 at 16:40
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In this answer I will carry out in detail the suggestions in Ian's comments.

Instead of constructing $\phi$, construct its Fourier transform $\Phi$. Let $$ b(\xi)=\begin{cases}c\,e^{-\tfrac{1}{1-\xi^2}} &\text{if }|\xi|\le1,\\0&\text{if }|\xi|>1, \end{cases}$$ where $c>0$ is chosen so that $\int_{-1}^1b(\xi)\,d\xi=1$. Next, let $\psi$ be the convolution of the characteristic function of the interval $[-\pi,\pi]$ with $\dfrac1\epsilon\,b\Bigl(\dfrac{\xi}{\epsilon}\Bigr)$. Then $\psi$ satisfies properties (1)-(4) as a function of $\xi$. Finally, observe that $\sum_{n\in\mathbb{Z}}\psi(\xi-2\pi n)$ is well defined, since only a finite number of summands are $\ne0$, and $>0$. Then define $$ \Phi(\xi)=\frac{\psi(\xi)}{\sum_{n\in\mathbb{Z}}\psi(\xi-2\pi n)} $$ and $\phi(x)$ as the inverse Fourier transform of $\Phi$.

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  • $\begingroup$ Thank you very much for your answer. There are two things I am still wondering about. (1) I am still wondering how to show $\psi$ is smooth? (2) $\Phi$ no longer has value $1$ in the interval $[-\pi+\epsilon, \pi - \epsilon]$. How can we get around that? Thank you ! $\endgroup$ – Johnny T. Dec 17 '15 at 6:22
  • $\begingroup$ 1) $\psi$ is the convolution of an integrable function with a $C^\infty$ function of compact support. By the properties of convolution, $\psi$ is smooth. 2) If $|\xi|<\pi-\epsilon$, then $\psi(\xi)=1$ and $\psi(\xi-2\,n\,\pi)=0$ for $n\ne0$. $\endgroup$ – Julián Aguirre Dec 17 '15 at 10:13
  • $\begingroup$ Thank you very much for the explanation! I get it now! $\endgroup$ – Johnny T. Dec 19 '15 at 18:10

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