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I am trying to find a function $f\in L_{PC}^{1}(-\infty ,\infty)\cap L_{PC}^{2}(-\infty ,\infty)$

such that $ \ \forall |\omega |<\pi \ \ \ \rightarrow \ \ \hat{f}(\omega )=\frac{1}{2 \pi}$

I have no restriction regarding $ |\omega |>\pi $

I know I cannot take the transform to be identically $0$ for $ |\omega |>\pi $, since then the transform will not be continuous and thus $f$ could not be in $L_{PC}^{1}(-\infty ,\infty)$.

The next step in my thinking was to take exponential decay, for example:

$$\hat{f}(\omega)=\left \{ \frac{1}{2\pi} \ \ \ |\omega|<\pi \ \ , \frac{e^{\pi-|\omega|}}{2\pi} \ \ |\omega|>\pi\right \} $$

However, to my dismay I found out that in this case, $f(x)$ is not absolutely integrable.

I think it is perhaps because the derivative of the transform has a jump discontinuities. I then proceeded to take Gaussians on each side to make sure it decays well to zero and the derivative of the transform is a continuous function. $$\hat{f}(\omega)=\left \{ \frac{1}{2\pi} \ \ \ |\omega|<\pi \ \ , \frac{e^{-(\omega-\pi)^2}}{2\pi} \ \ \omega>\pi \ , \ \frac{e^{-(\omega+\pi)^2}}{2\pi} \ \ \omega<-\pi \right \} $$ Unfortunately, computing $f(x)$ (via the inverse transform formula) seemed to be too complicated (it involved the error function).

I would very much like to hear your input about my thinking, if it's in the right direction or am I off entirely?

In addition, is there some general rule about how smooth does the transform have to be in order for the function to be absolutely and square integrable?

Thank you in advance!

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  • $\begingroup$ Piece-wise continuous $\endgroup$ – zokomoko Jul 9 '16 at 7:58
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I define $\hat f$ by:

$$\hat f (t) = \frac1{\sqrt {2\pi}} \int_{\Bbb R} f(\xi) e^{-i\xi t} d\xi$$

Let:

$$f(x) = \frac1{\pi \sqrt{2\pi}}\frac{\sin((\pi + 1) x) \sin x}{x^2}$$

Obviously, $f\in L^1_{PC} \cap L^2_{PC}$ and check that:

$$\hat f(t) = \frac{1}{4\pi} \chi_{[-a,a]} \star \chi_{[-1,-1]}$$

where $a = 1 + \pi$. This $\hat f$ satisfies the requirement.

What I did is not very special: I remember that $\chi_{[-a,a]} \star \chi_{[-1, 1]}$ is an isosceles trapezoid on $[-a - 1, a +1]$ and zero elsewhere, and has constant value $2$ on $[- a+1, a -1]$. I modified it by multiplying by that factor which appears in $\hat f$, then took the FT to get my $f$.

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  • $\begingroup$ thank you! you helped me a great deal! $\endgroup$ – zokomoko Jul 10 '16 at 9:03

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