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
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$1/\sqrt{x}$ is not in $L^\infty$ on $[0,\infty)$. – mrf Sep 23 '16 at 11:50
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Sorry prof. Interval is on [1,\infty) – user352724 Sep 23 '16 at 11:53
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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 $L^\infty ([1, \infty))$.
Alex M.
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Indeed, it's worth noting that this rule extends to $\mathbb{R}^n$: $x^{\alpha} \in L^1(\mathbb{R}^n)$ if $\alpha > -n$. – NoseKnowsAll Nov 29 '16 at 05:00
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1@NoseKnowsAll: Not to $\Bbb R^n$, but rather to the unit ball or the unit cube (or, probably, any other compact neighbourhood of $0$) in $\Bbb R^n$. And the function becomes $\frac 1 {| x | ^\alpha}$. – Alex M. Nov 29 '16 at 14:45