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.


3 Answers 3


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. $\square$


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,

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


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

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 $\int_0 ^\infty f(x) x^\alpha dx <\infty$.

  • $\begingroup$ Seems we had a very similar idea 😄 $\endgroup$ Jan 27, 2020 at 22:56

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, $\lim_{x\to \infty} f(x) x^{\alpha+1} = 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.