Let $\;f\;: \mathbb R^m \rightarrow [0, \infty)\;$ be a continuous function and $\;g:[M,+\infty)\rightarrow \mathbb R\;$ a non-negative $\;L_{loc}^1\;$ function, where $\;M \gt 0\;$ such that $\; \int_{M}^{\infty} g(s) \;ds=+\infty\;$.
If this holds:
$\;\sqrt {f(x)} \ge g(\vert x \vert )\;\;\forall x:\;\vert x \vert \ge M\;$ (*),
then one can say $\;f\;$ decays slow to $\;0\;$ at infinity.
Why is this true?
I thought of setting $\;g(x)=\frac {1}{\sqrt x}\;$ to (*) . Then I get $\;f(x) \ge \frac {1}{\vert x \vert}\;\;\forall x:\;\vert x \vert \ge M\; \Rightarrow \int_{M}^{\infty} f(x)\;dx \ge \int_{M}^{\infty} \frac {1}{\vert x \vert}\;dx =+\infty $
However it's still unclear to me how this connects to the behaviour of $\;f\;$ at infinity. Does the fact that $\;\int_{M}^{\infty} f(x)\;dx \ge +\infty\;$ imply somehow $\;f\;$ slow decays to $\;0\;$?
I 'm really new to fast or slow decay of a function, so I apologize in advance if my question is trivial or quite elementary. Any help would be valuable.
Thanks!