Proving that $\widehat{g*f}\in L^1(\mathbb{R})$ with $f(t) = \frac{1}{\sqrt{1 + t^4}}$ and $g(t) = e^{-|t|}$

Question

The question is as follows:

Verify whether the Fourier of the convolution: $f * g$ belongs to $L^1(\mathbb{R})$? Where $$f(t) = \frac{1}{\sqrt{1 + t^4}}~~and ~~~~g(t) = e^{-|t|}$$

$\textbf{Efforts:}$

As we know $L^1(\mathbb{R}) $ is closed under convolution. So is it enough to show that $f(t)$ and $g(t)$ are in $L^1(\mathbb{R})$ for to see that their convolution is also in $L^1(\mathbb{R})$? Can you please let me know what conditions has to satisfy for the convolution be in $L^1(\mathbb{R})$?

Answer 1

Indeed, see here \begin{split} \widehat{g}(x)= \int_{-\infty}^{\infty}e^{-|t|}e^{-ix t}\,dt &=&\int_{-\infty}^{0}e^{t}e^{-ix t}\,dt+\int_{0}^{\infty}e^{-t}e^{-ix t}\,dt\\ &=&\left[ \frac{e^{(1-ix)t}}{1-ix} \right]_{-\infty}^0-\left[\frac{e^{-(1+ix)t}}{1+ix} \right]_{0}^{\infty}\\ &=&\frac{1}{1-ix}+\frac{1}{1+ix}\\ &=&\color{blue}{\frac{2}{x^2+1}\in L^1(\Bbb R)}. \end{split}

Therefore, we have that $\widehat{g}\in L^1$ and we know that $\widehat{f} \in L^\infty$ (Since $f\in L^1$ ) hence $$\color{red}{\widehat{f*g}=\widehat{f}\displaystyle\widehat{g} \in L^1}$$

Proof of the claim above $$|\widehat{f}(a)|=\left|\int f(x)e^{-ixa}\, dx\right| \le \int \left|f(x)\, \right|dx =\|f\|_1 $$

that is $$\|\widehat{f}\|_{\infty}\le \|f\|_1 $$