I very much dislike the "Big Oh" notation. It just doesn't stick in my mind. Suppose $f$ is a continuous function and $f \in \text{O}( 1/|x|^{1+\epsilon})$ when $|x| \rightarrow \infty$ and for $0< \epsilon < 1$. Does this mean that $$ \int_{-\infty}^\infty |f(x)|\cdot |x|^\epsilon \; dx < \infty ?$$


No. This only gives the trivial upper bound

$$\int_1^{\infty} |f(x)| |x|^{\epsilon} \, dx \le \int_1^{\infty} \frac{C}{|x|} \, dx = \infty$$

for some constant $C$.

  • $\begingroup$ Example. If $f(x)=|x|^{-1-\epsilon}$ when $|x|\geq 1.$ $\endgroup$ – DanielWainfleet Apr 9 '16 at 19:40

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