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Suppose that $f$ is a decreasing continuous function on $[0, \infty)$. And the integral of $f(x)/\sqrt{x}$ on $[0, \infty)$ exists.
Prove that $\lim_{x\rightarrow \infty}\sqrt{x}f(x)=0$.
My work. I think we should prove the limit exists firstly. Then everything would be quite easy.
Suppose that $\exists \epsilon>0$, $\forall N>0$, $\exists x_N>N$, s.t $\sqrt{x_N}f(x_N)>\epsilon$. We would find that $f(x_N)/\sqrt{x_N}>\epsilon/x_N$.
But I can't go any further. Any one has some better idea?
Hint. Since $f$ is decreasing in $[0, \infty)$, then also $f(x)/\sqrt{x}$ is decreasing in $(0, \infty)$: if $0<x<y$ then $$\frac{f(x)}{\sqrt{x}}\geq \frac{f(y)} {\sqrt{x}}\geq\frac{f(y)}{\sqrt{y}}.$$ Hence, for $x>0$, $$0\leq f(x)\sqrt{x}=2\frac{f(x)}{\sqrt{x}} (x - x/2) \leq 2\int_{x/2}^{x} \frac{f(t)}{\sqrt{t}} \, dt.$$ Now use the fact that the integral $\int_{1}^{\infty}\frac{f(t)}{\sqrt{t}}\,dt$ is finite to show that the r.h.s. goes to zero as $x\to+\infty$.
P.S. Note that $f(x)$ is non negative. If $f(x_0)<0$ for some $x_0>0$ then $-f(x)\geq -f(x_0)> 0$ and $$-\int_{x_0}^{\infty}\frac{f(t)}{\sqrt{t}}\,dt\geq (-f(x_0))\int_{x_0}^{\infty}\frac{1}{\sqrt{t}}\,dt=+\infty$$ which contradicts the fact that $\int_{1}^{\infty}\frac{f(t)}{\sqrt{t}}\,dt$ is finite.
