# If $f:[0,\infty)\to[0,\infty)$ is continuous & decreasing and $α>0$ s.t. $\int_{0}^\infty x^α f(x)dx<\infty$, $\lim_{x\to\infty}f(x)x^{α+1}=0$

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.

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,

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

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

• Seems we had a very similar idea 😄 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, $$\lim_{x\to \infty} f(x) x^{\alpha+1} = 0$$.