What if we define function of moderate decrease as function satisfying $|f|\le\frac{A}{x^2}$ In Stein's Fourier Analysis, he defines the a continuous function $f$ as of moderate decrease if there exists $A>0$ such that
$$|f(x)| \le \frac{A}{1+x^2} \forall x\in\mathbb{R}$$
I am wondering what if we just define it as $|f(x)|\le \frac{A}{x^2}$? I am asking this because it seems like we don't need that $1$ when showing "whenever $f$ belongs to $\mathcal{M}(\mathbb{R}),$ then we can define$\int_{-\infty}^{\infty} f(x)dx $.
The proof basically argues that the sequence $I_n := \int_{-N}^N f(x) dx$ is Cauchy. And the derivation goes like
$$|I_M - I_N | \le | \int_{N\le |n| \le M} f(x) dx |$$
$$ \le \int_{N\le |n| \le M} |f(x)| dx$$
$$\le \int_{N\le |n| \le M} \frac{A}{x^2} dx$$
$$ \le A \int_{N\le |n| \le M} \frac{dx}{x^2}$$
So I feel like the $1$ in the definition probably does not matter? But there should be a reason...


I am wondering what if we just define it as $|f(x)|\le \frac{A}{x^2}$?

The function $ x \mapsto \frac1{x^2}$ is not in $L^1(\mathbb{R})$ whereas $ x \mapsto \frac1{1+x^2}$ is in $L^1(\mathbb{R})$.
