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Let $f: [0,\infty)\rightarrow [0,\infty)$ be a continuous and decreasing function. Suppose that exists an $\alpha>0$ such that $\int_{0}^{\infty} x^\alpha f(x) dx < \infty$. Prove that $\lim_{x\rightarrow \infty} f(x)x^{\alpha+1}=0$.

First I tried to integrate using parts formula because I thought that was a good way for arriving to the limit. And then I tried using the definition of convergence but I didn't know how to finish.

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3 Answers 3

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Indeed suppose that the statement $\lim_{x\rightarrow \infty} f(x)x^{\alpha+1}=0$ is false. Then, by definition of the limit, there exists a $\varepsilon>0$ such that for every $X\in[0,\infty[$, there exists a $x\geq X$ such that $f(x)x^{\alpha+1}\geq\varepsilon$.

Hence we can construct a sequence $x_1,x_2,x_3,\dots$ in $[0,\infty[$ satisfying:

  • $f(x_i) x_i^{\alpha+1}\geq\varepsilon$ for all $i$, and
  • $\frac{x_{i}}{x_{i-1}}\geq 2^i$ for all $i$.

Now comes the main idea: Since $f$ is decreasing, we have $f(x)\geq\frac{\varepsilon}{x_i^{\alpha+1}}$ for all $i$ and $x\le x_i$.

Hence, for all $i\geq 2$, $$\int_{x_{i-1}}^{x_i} f(x) x^\alpha\,\mathrm dx\geq\varepsilon\int_{x_{i-1}}^{x_i} \frac{x^\alpha}{x_i^{\alpha+1}}\,\mathrm dx=\frac\varepsilon{\alpha+1}-\varepsilon\left(\frac{x_{i-1}}{x_i}\right)^\alpha\geq\frac{\varepsilon}{\alpha+1}-\frac{\varepsilon}{2^i}.$$

It follows that $$\int_0^\infty f(x) x^\alpha\,\mathrm dx\geq \sum_{i=2}^\infty \int_{x_{i-1}}^{x_i} f(x) x^\alpha\,\mathrm dx\geq\sum_{i=2}^\infty\frac{\varepsilon}{\alpha+1}-\frac{\varepsilon}{2^i},$$ but the last sum is clearly divergent. Contradiction. $\square$

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We argue by contradiction. If not, then

$$\lim _{x\to \infty} f(x) x^{\alpha +1} =0$$

is false, and thus there is an $\epsilon_0 >0$ and a sequence $\{s_n \}$ of positive real numbers so that

$$ f(s_n) s_n^{\alpha +1} \ge \epsilon _0$$

for all $n\in \mathbb N$.

Now since $\int_0^\infty f(x) x^\alpha dx <\infty$, there is $M>0$ so that

$$ \int_M ^y f(x) x^\alpha dx < \frac{\epsilon_0}{2(\alpha +1)}$$

for all $y>M$. Since $\{s_n\}$ converges to infinity, for all $x>M$, there is $s_n$ so that $s_n >x$.

since $f$ is decreasing,

$$ \int_x^{s_n} f(t) t^\alpha dt \ge \int_x^{s_n} f(s_n) t^\alpha dt= \frac{f(s_n)}{\alpha +1} ( s_n^{\alpha +1} - x^{\alpha +1})\ge \frac{1}{\alpha +1}( f(s_n)s_n^{\alpha +1} - f(x) x^{\alpha +1}). $$

So

$$\epsilon_0 \le f(s_n) s_n^{\alpha +1} \le(\alpha +1) \int_x^{s_n} f(t) t^\alpha dt + f(x) x^{\alpha+1}\le \epsilon_0/2 + f(x) x^{\alpha+1}$$

which implies $f(x) x^{\alpha +1} \ge \epsilon_0/2$ for all $x>M$.

But then

$$ \int_M ^y f(x) x^\alpha dx\ge \frac{\epsilon_0}{2} \int_M^y \frac{1}{x} dx = \frac{\epsilon}{2} (\ln y - \ln M) $$

which is unbounded as $y \to +\infty$. This is a contradiction to the assumption that $\int_0 ^\infty f(x) x^\alpha dx <\infty$.

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  • $\begingroup$ Seems we had a very similar idea 😄 $\endgroup$ Jan 27, 2020 at 22:56
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Since $f$ is a decreasing function then $$0\leq \frac{f(x) x^{\alpha + 1}}{\alpha + 1} \leq f(x) \int_0^x t^\alpha dt \leq \int_0^x f(t) t^\alpha dt \leq \int_0^\infty f(t) t^\alpha dt,$$ which implies $$\lim_{x\to \infty} f(x) = 0.$$ Let $A >0$ be an arbitrary number and $x > A$. Since $f$ is decreasing then we have $$\int_A^\infty f(t) t^\alpha dt \geq \int_A^x f(t) t^\alpha dt \geq f(x)\int_A^x t^\alpha dt = f(x) \frac{x^{\alpha+1} -A^{\alpha+1}}{\alpha+ 1},$$ or equivalently $$f(x) x^{\alpha+1} \leq f(x) A^{\alpha+1} + (\alpha+1)\int_A^\infty f(t) t^\alpha dt.$$ Letting $x\to \infty$, we get $$\limsup_{x\to \infty} f(x) x^{\alpha+1} \leq (\alpha+1)\int_A^\infty f(t) t^\alpha dt$$ for any $A >0$. Letting $A \to \infty$ and using $\int_0^\infty f(t) t^\alpha dt < \infty$ we get $$\limsup_{x\to \infty} f(x) x^{\alpha+1} \leq 0.$$ Evidently, we have $$\liminf_{x\to \infty} f(x) x^{\alpha+1} \geq 0.$$ Hence, $\lim_{x\to \infty} f(x) x^{\alpha+1} = 0$.

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