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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...

3x89g2
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1 Answers1

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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 $L^1(\mathbb{R})$.

Olivier Oloa
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