an inequality on convolution

Question

Let $f(x)=A(1+|x|)^{-M}$ and $g(x)=B(1+|x|)^{-N}$ be two functions on $\mathbb{R}$. Here $M\ge N>1$. I'm trying to prove the following inequality about convolution $$ |(f*g)(x)|\le ABC(1+|x|)^N, $$ where $C$ depends on $M$ and $N$. I tried using $(1+|x-y|)^{-1}\le (1+|y|)(1+|x|)^{-1}$ to get rid of $(1+|x-y|)$ in the definition of convolution, but I could not get a finite integral to be represented by $C$. Any ideas?

Answer 1

It follows from the triangle inequality that

$$(1+\left|y\right|)\leq(1+\left|x-y\right|)(1+\left|x\right|)\Leftrightarrow (1+\left|x-y\right|)^{-1}\leq (1+\left|x\right|)(1+\left|y\right|)^{-1}$$

Whence,

$$\begin{array}{lcl}\displaystyle\int_{\mathbb{R}}f(y)g(x-y)dy&\leq&\displaystyle\int_{\mathbb{R}}A(1+\left|y\right|)^{-M}\cdot B(1+\left|x\right|)^{N}(1+\left|y\right|)^{-N}dy\\[2 em] &=&\displaystyle AB(1+\left|x\right|)^{N}\underbrace{\int_{\mathbb{R}}(1+\left|y\right|)^{-M-N}dy}_{:=C}\end{array}$$

Since $M+N\geq 2$, the function $(1+\left|y\right|)^{-M-N}$ is integrable; therefore $C$ is finite constant depending on $M,N$.

You can actually do much better and show that

$$\left|(f\ast g)(x)\right|\leq ABC\left(1+\left|x\right|\right)^{-N},$$

for some constant $C>0$. Indeed, for $x\in\mathbb{R}$, split $\mathbb{R}$ into the sets $\left\{\left|x-y\right|\geq\left|x\right|/2\right\}$ and $\left\{\left|x-y\right|<\left|x\right|/2\right\}$. Observe that

$$\begin{array}{lcl}\displaystyle\int_{\left|x-y\right|\geq\frac{\left|x\right|}{2}}(1+\left|x-y\right|)^{-M}(1+\left|y\right|)^{-N}dy&\leq&\displaystyle\int_{\left|x-y\right|\geq\frac{\left|x\right|}{2}}(1+\frac{\left|x\right|}{2})^{-M}(1+\left|y\right|)^{-N}dy\\[2 em]&\leq&\displaystyle 2^{M}(1+\left|x\right|)^{-M}\int_{\mathbb{R}}(1+\left|y\right|)^{-N}dy \end{array}$$

If $\left|x-y\right|\leq\left|x\right|/2$, then it follows from the reverse triangle inequality that $\left|y\right|>\left|x\right|/2$. Whence,

$$\begin{array}{lcl}\displaystyle\int_{\left|x-y\right|\leq\frac{\left|x\right|}{2}}(1+\left|x-y\right|)^{-M}(1+\left|y\right|)^{-N}dy&\leq&\displaystyle\int_{\left|x-y\right|\leq\frac{\left|x\right|}{2}}(1+\left|x-y\right|)^{-M}(1+\frac{\left|x\right|}{2})^{-N}dy\\[2 em] &\leq&\displaystyle2^{N}(1+\left|x\right|)^{-N}\int_{\mathbb{R}}(1+\left|x-y\right|)^{-M}dy\end{array}$$

Taking

$$C:=\min\left\{2^{M}\int_{\mathbb{R}}(1+\left|y\right|)^{-N}dy,2^{N}\int_{\mathbb{R}}(1+\left|y\right|)^{-M}dy\right\}$$

completes the proof.