Q:

Necessary condition for an improper integral to converge

I am working on this problem from a past examination:

Let $f:[0,\infty)\rightarrow\mathbb R$ be a continuous, non-negative and non-increasing function such that the improper integral $\int_0^\infty (f(x)/\sqrt{x})\ dx$ converges.  Show that $\lim_{x\rightarrow\infty}f(x)\sqrt x=0$.  Also, prove that $0<\forall \epsilon<1$, $$\lim_{x\rightarrow\infty}\int_{\epsilon x}^x\frac{f(y)}{\sqrt{x-y}}dy = 0.$$

I tried to bound the order of growth of $f$ to show the first assertion, but in vain.
I would be grateful if you could provide a clue (not necessarily a complete proof).

A:

Here are some ideas that might help:
If $f(b)\sqrt{b}\ge\varepsilon$, then $f(x)\ge\varepsilon/\sqrt{b}$ for all $x\le b$.
The assumption on the integral implies that $$\lim_{b\to\infty}\int_{b/2}^b\frac{f(x)}{\sqrt{x}}\,dx=0.$$

