# Existence of integral implies the existence of a limit

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?

• By $f$ being "non-increasing", do you mean "$f$ is not monotonically increasing", i.e. it has at least one place where it decreases, or "$f$ never increases"? I would call that last one a decreasing function. Jul 19, 2017 at 6:34
• @Arthur non-increasing means that if $a>b$ then $f(a)\le f(b)$ that is it never increases. Jul 19, 2017 at 6:35
• @skyking To me, a function being non-increasing is a function which is not an increasing function. What you have there is exactly what a decreasing function is defined as. Jul 19, 2017 at 6:36
• @Arthur Forgot it and try to focus on the question. Jul 19, 2017 at 6:39
• For the benefit of everybody... Jul 19, 2017 at 6:41

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.