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 $$\lim_{b\to\infty}\int_{b/2}^b\frac{f(x)}{\sqrt{x}}\,dx=0.$$
A: I would try to absorb $\sqrt{x}$ in the integral, into the differential $dx$, by using the relation $d(x^{3/2})=\frac32\sqrt{x}dx$, and then use the fact that the function remains monotone after the change of variable.
A: Based on the accepted answer, I provide a complete solution to the problem.
The first proposition follows from the fact that
$$
0 \le \frac{f(x)\sqrt{x}}{2} = \frac{xf(x)}{2\sqrt{x}}\le\int_{x/2}^x\frac{f(\xi)}{\sqrt \xi}d\xi\longrightarrow 0\ (x \longrightarrow \infty).
$$
The inequality on the right follows from the monotonicity of $f$ and thus $\frac{f(\xi)}{\sqrt \xi}$.
Next,
$$
0 \le \int_{\epsilon x}^x\frac{f(y)}{\sqrt{x-y}}dy \le f(x)\int\frac{dy}{x-y} = 2f(x)\left[\sqrt{x - y}\right]^{y=\epsilon x}_{y=x} = 2f(x)\sqrt{(1-\epsilon) x} \le 2f((1-\epsilon) x)\sqrt{(1-\epsilon) x} \rightarrow 0\ (x\rightarrow \infty).
$$
