$f_a\chi_{(0,1)}$ is in $L^p$ iff $p<a^{-1}$

Question

From Folland "Real Analysis":

"It is instructive to consider following simple examples on $(0, \infty)$ wtih Lebesgue measure. Let $f_a(x) = x^{-a}$, where $a>0.$ Elementary calculus shows that $f_a\chi_{(0,1)}\in L^p$ iff $p<a^{-1}$, and $f_a\chi_{(1, \infty)} \in L^P$ iff $p > a^{-1}$."

I don't quite understand what the author did here.

I know that $f \in L^p$ if $|f|^p$ is Lebesgue integrable. And $$\int|x^{-a}\chi_{(0,1)}|^p = \int|x^{-a}|^p\chi_{(0,1)}.$$

So, for example, if $p = 1/a^2<1/a$, $$\int|x^{-1/a}|\chi_{(0,1)},$$ and this is integrable. But if $p=2/a\ge 1/a,$ $$\int|x^{-2}|\chi_{(0,1)},$$ and this is not integrable.

Could you explain why?

Answer 1

Your computations just prove Folland's claim in the special cases $p = 1/a^2$ (if $a > 1$ otherwise $1/a^2 \geq 1/a$) and $p = 2/a$. The proof of the claim for general $a > 0$ is not much harder. So let us check for which $a >0$ the function $f_a \chi_{(0,1)} \in L^p(0, \infty)$: $$ \int_0^\infty \lvert f_a(x) \chi_{(0,1)}(x)\rvert^p \, \text{d} x = \int_0^1 x^{-ap} \, \text{d}x = \frac{1}{1-ap}\Big[ x^{1-ap} \Big]_{x=0}^{x=1} = \begin{cases} \frac{1}{1-ap}& ap < 1 \\ + \infty & ap > 1 \end{cases}\, . $$ So $f_a \chi_{(0,1)} $ is in $L^p(0, \infty)$ if and only if $ap < 1$, which is the first part of the claim. The proof of the second part is similar.

Answer 2

Take $f_a \chi_{(0,1)}$ as example. Since $f_a \chi _{(0,1)}$ is nonnegative and measurable, by [Beppo Levi's] monotonic convergence theorem we have \begin{align*} &\phantom{==}(\mathrm L)\int x^{-ap}\chi _{(0,1)}(x)\mathrm dx \\ &= \lim_n (\mathrm L)\int x^{-ap}\chi_{[1/n, 1)} \mathrm dx \\ &=\lim_n (\mathrm R)\int_{1/n}^1 x^{-ap}\mathrm dx \\ &= \begin{cases} \lim_n \dfrac {1 - (1/n)^{1-ap}}{1-ap}, & ap\neq 1,\\ \lim_n -\log(1/n) = +\infty, & ap = 1, \end{cases}\\ &= \begin{cases} +\infty, & ap\geqslant 1,\\ \dfrac 1 {1-ap} < +\infty, & ap < 1, \end{cases} \end{align*} hence the claim.