If $0\leq \alpha\leq1$, then $f(x)=x^{-\alpha}$ is not $L^{1}$ on $[1,\infty)$. But $f(x)$ is $L^{\infty}$ on $[1,\infty)$. How to show this

Question

If $0\leq \alpha\leq1$, then $f(x)=x^{-\alpha}$ is not $L^{1}$ on $[1,\infty)$. But $f(x)$ is $L^{\infty}$ on $[1,\infty)$. How to show this

Answer 1

Well, notice that if $0 \le \alpha < 1$

$$\int \limits _1 ^\infty \frac 1 {x^\alpha} \ \Bbb d x = \frac {x^{-\alpha + 1}} {-\alpha + 1} \Bigg| _1 ^\infty = \infty - \frac 1 {-\alpha + 1} = \infty$$

therefore $\dfrac 1 {x^\alpha}$ is not in $L^1 ([1, \infty))$.

If $\alpha = 1$, then

$$\int \limits _1 ^\infty \frac 1 x \ \Bbb d x = \ln x \Big| _1 ^\infty = \infty ,$$

so $\dfrac 1 x$ is again not in $L^1 ([1, \infty))$.

---

On the other hand, if $0 \le \alpha \le 1$, then $\dfrac 1 {x^\alpha}$ is decreasing, so $\dfrac 1 {x^\alpha} \le \dfrac 1 1 = 1$, so $\dfrac 1 {x^\alpha}$ is bounded, therefore in $L^\infty ([1, \infty))$.