Relationship between integrability of function and smoothness of its fourier transform

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?

• Piece-wise continuous – zokomoko Jul 9 '16 at 7:58

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 $f$.

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