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I need to prove that, let $f : \mathbb{R} \to \mathbb{R} : f(t) = 1/(1+t^2)$, $\hat{f}$ is square integrable. Now, I know that Fouriertransforms are continuous, thus $\hat{f}$ will surely be integrable over any closed, bounded interval. So I need to check the behaviour for $t \to \infty$. I also know that the smoother $f$ is, the quicker $\hat{f}$ will go to zero for $t \to \infty$. So I need to check the smoothness of $f$? How do I do this, or what key propositions are there to help me with this?

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  • $\begingroup$ You do this by checking if you can differentiate $f$.......? $\endgroup$ Jan 22, 2022 at 14:03
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    $\begingroup$ $f$ belongs to $L^2$, and the Fourier transform is an isomorphism on $L^2$, so $\hat{f}\in L^2$ as well. $\endgroup$
    – peek-a-boo
    Jan 22, 2022 at 14:03
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    $\begingroup$ $\hat{f}(\omega)=\pi e^{-|\omega|}$ $\endgroup$
    – wimi
    Jan 22, 2022 at 14:20

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Yes. For $|t|>1$, $f(t)< \frac{1}{t^2}$, which you can easily show is square integrable.

Apply plancherel's theorem to conclude that $\hat{f}$ is also square integrable.

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Since $\forall t \in \mathbb{R}: \frac {1}{1+t^2} \le 1,$

$$\int_{- \infty} ^ {\infty} \left(\frac {1}{1+t^2} \right)^2 dt \le \int_{- \infty} ^ {\infty} \frac {1}{1+t^2} dt = \arctan(t)|_{- \infty}^{\infty} = \pi.$$

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