why the convolution of two functions of moderate decrease is again function of moderate decrease?

Question

I need to prove that a convolution of 2 functions of moderate decrease is a function of moderate decrease. I tried to split the integral into two integrals but I couldn't manage to bound any one of them by $\frac{A}{x^2+1}$.

Thanks

Answer 1

Take $z = x - y/2$. $$ \begin{align*}\int_{-\infty}^{\infty} \frac{1}{1 + x^2} \cdot \frac{1}{1 + (y-x)^2} \mathrm{d}x &= 2 \int_0^\infty \frac{1}{1 + (y/2 + z)^2} \cdot \frac{1}{1 + (y/2 - z)^2} \mathrm{d}z \\ & \leq \frac{2}{1 + (y/2)^2} \int_0^\infty \frac{1}{1 + (y/2 - z)^2} \mathrm{d}z \\ & \leq \frac{2}{1 + (y/2)^2} \int_{-\infty}^\infty \frac{1}{1 + (y/2 - z)^2} \mathrm{d}z \\ & = \frac{2}{1 + (y/2)^2} \int_{-\infty}^\infty \frac{1}{1 + w^2} \mathrm{d}w \end{align*}$$

Lastly observe $$ \frac{2}{1 + (y/2)^2} \leq \frac{2}{1/4 + (y/2)^2} \leq \frac{8}{1 + y^2} $$

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The power $2$ in $(1 + x^2)^{-1}$ is not special. For any power $1+\epsilon$ where $\epsilon > 0$ you get a similar result.