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Let $f$ nonnegative continuous decreasing function on $[0,\infty)$. Suppose $\frac{f(x)}{\sqrt{x}}$ is integrable, then $\sqrt{x}f(x) \to 0$ as $x\to \infty$

I tried to prove $\sqrt{x}f(x)$ is Cauchy sequence, but I cannot to do. And I don't know how to use continuity.

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2 Answers 2

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Let's just shift the problem to make it easier:

Let $f: \mathbb{R}^+ \to \mathbb{R}$ be a continuous, decreasing and integrable (on $\mathbb{R}^+$) function. Prove that $xf(x)\to 0$ as $x\to \infty$.

First step: Notice that $f\to 0$ as $x\to \infty$ (fairly well-known fact but still)

Then $f$ has to be nonnegative (we did not need it as an hypothesis!) because it decreases towards $0$.

Second step:

$$ 0\leq \frac{x}{2}f(x) \leq \int_{x/2}^x f(t) dt \underset{x\to \infty}{\to} 0 $$

Because $\int_{x/2}^x f(t) dt = \int_0^x f(t) dt - \int_0^{x/2} f(t) dt $ and both integrals converge to the same limit.

Hence the result.

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  • $\begingroup$ Would those who downvoted care to explain? Thanks in advance. $\endgroup$ Jun 18, 2018 at 14:01
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Take $0<u<v$. You have for $x \in [u,v]$ $\frac{f(u)}{\sqrt{x}} \le \frac{f(x)}{\sqrt{x}} \le \frac{f(v)}{\sqrt{x}} $

Hence

$$0 \le 2\left(\sqrt{v}-\sqrt{u}\right) f(u)\le \int_u^v F(t) \ dt\le 2\left(\sqrt{v}-\sqrt{u}\right) f(v)$$ where $F(x) = \frac{f(x)}{\sqrt{x}}$

Now take $u=x$ and $v=4x$ you get $$0 \le \sqrt{x}f(x) \le \frac{1}{2}\int_{x}^{4x} F(t) \ dt \underset{x\to \infty}{\to} 0$$

using Cauchy criteria for a converging integral. Hence $\lim\limits_{x \to \infty} \sqrt{x}f(x) = 0$.

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