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

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