# If $f : \mathbb{R} \to \mathbb{R} : f(t) = \frac{1}{1+t^2}$, is $\hat{f}$ square integrable?

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?

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

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.
Since $$\forall t \in \mathbb{R}: \frac {1}{1+t^2} \le 1,$$
$$\int_{- \infty} ^ {\infty} \left(\frac {1}{1+t^2} \right)^2 dt \le \int_{- \infty} ^ {\infty} \frac {1}{1+t^2} dt = \arctan(t)|_{- \infty}^{\infty} = \pi.$$