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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

user148728
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  • What is your precise definition for moderate decrease? There are several (equivalent) definitions, but it will be easier to write an answer using the one you are familiar with instead of a different one. – Willie Wong Jun 15 '15 at 11:34
  • definition of moderate decrease function- A function f defing on R is said to be moderate decrease if f is continuous and there exists a constant A>0 so that A/(x^2+1) – user148728 Jun 16 '15 at 11:44

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

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

Willie Wong
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    $\frac{1}{1+(y/2+z)^2} \leq \frac{1}{1+(y/2)^2}$ only if $y$ is positive. If it's negative, use $\frac{1}{1+(y/2-z)^2} \leq \frac{1}{1+(y/2)^2}$. You can then proceed with the same proof. – atreju Oct 24 '19 at 22:27