I am trying to conclude an $\varepsilon,\delta$ proof. I have a non-decreasing function $f(r) \geq 0$ that converges to $0$ when $r \to 0 $ $(r >0)$. I want to find a sequence $r_n$ such thant $n r_n^2 f(r_n) \to 0$ and $n r_n^2 \to \infty$ when $n \to \infty$. When $f$ is smooth we can choose $r_n = \frac{\ln(n)}{n^{1/2}}$.
What about the general case ? My idea was to regularize $f$ by a function $f^+$ such that $f^+(r)$ converges to $0$ when $r \to 0$ and $f \leq f^+$.
Is it possible to do that ?
Thank you.
Edit :
If it can help, the function $f$ has the following expression :
$f(r) = \Bbb E\left(\sup\limits_{|h| \leq r} g(h)\right)$ where $g(h)$ is a positive random function that satisfies $g(h) \to 0$ when $h \to 0$ and $g(h) \leq g^+$ with $g^+$ integrable.
Edit 2 :
In fact we can't find in general such a $f^+$. Take $f(r) = \sqrt{r}$. Then $\frac{f(r)}{r} = \frac1{\sqrt{r}} \leq \frac{f^+(r) - f^+(0)}{r}$.
But, I can't prove or disprove my first question. It can be reformulated as : find a sequence $a_n$ such that $a_n \to \infty$ and $a_n f(\frac{a_n}{\sqrt{n}}) \to 0$.
I can prove it when $\limsup_{r \to 0} f(r)/r < \infty$. But, the example $f(r) = \sqrt{r}$ show that we should be able to find $a_n$ in a more general setting (indeed we can take $a_n = \ln(n)$).