Question on $L_p$ spaces involving $\lambda^n$-measure on $\mathbb{R}^n$ Q/ Consider $L_p=L_p(\lambda^n)$ with the Lebesgue measure on $\mathbb{R}^n$ and $1\leq p<\infty$. Let $f_0=|x|^{-\alpha}$ for $|x|<1$ and $0$ otherwise.
Show $f_0\in L_p$ iff $p\alpha < n$.
Now first I have shown that $f_0$ is $\lambda^n$-measurable for all n so the question is equivalent to (by the definition of an $L_p$ space) showing
$||f_0||_p=\big (\int_{\mathbb{R^n}}f_0\;d(\lambda^n)\big )^{\frac{1}{p}}<\infty$ iff $p\alpha < n$.
I was thinking the way to proceed would be to look at the integral explicitly and try and evaluate it or potentially find some useful bounds;
So $\big (\int_{\mathbb{R^n}}f_0\;d(\lambda^n)\big )^{\frac{1}{p}}=\big (\int_{\{x\;:\;|x|<1\}}|x|^{-\alpha}\;d(\lambda^n)\big )^{\frac{1}{p}}$
but then I am now stuck, I don't know how to compute this integral and can't find any meaningful bounds. Any help would be appreciated.
Thanks
 A: You've misstated the definition of $L^p$. You should have
$$ \|f_0\|^p = \left( \int_{\mathbb R^n} f_0^p \, dx \right)^{1/p}.$$ Where $dx$ represents integration with respect to Lebesgue measure.
One way to evaluate the integral is to integrate $f_0^p$ over concentric rings centered at $0$. Given $0 < t < 1$ you can write
$$
\{x : |x| < 1 \} = \bigcup_{k=0}^\infty \{x : t^{k+1} \le |x| < t^k\}.$$
Since this union is disjoint you have that
$$
\int_{\{|x| < 1\}} f_0^p \, dx = \int_{\{|x| < 1\}} |x|^{-\alpha p} \, dx = \sum_{k=0}^\infty \int_{\{ t^{k+1} \le |x| < t^k\}} |x|^{-\alpha p} \, dx.$$
Let $\omega$ denote the volume of the unit ball in $\mathbb R^n$. The Lebesgue measure of the ring $\{t^{k+1} \le |x| < t^k\}$ is then given by $\omega(t^{nk} - t^{n(k+1)})$ so that
$$
\omega t^{-\alpha p k}(t^{nk} - t^{n(k+1)}) \le  \int_{\{ t^{k+1} \le |x| < t^k\}} |x|^{-\alpha p} \, dx \le \omega t^{-\alpha p (k+1)}(t^{nk} - t^{n(k+1)}).$$
Cleaned up a bit, this gives you
$$\omega t^{(n-\alpha p)k}(1-t^n) \le  \int_{\{ t^{k+1} \le |x| < t^k\}} |x|^{-\alpha p} \, dx \le \omega t^{(n-\alpha p)k}(1-t^n)t^{-\alpha p}.$$
If $t^{n - \alpha p} < 1$ (which will be the case if and only if $n - \alpha p > 0$) you can sum over $k$ to yield
$$\omega (1-t^n) \frac{1}{1-t^{n-\alpha p}} \le \int_{\{|x| < 1\}} |x|^{-\alpha p} \, dx \le \omega (1-t^n)t^{-\alpha p} \frac{1}{1-t^{n-\alpha p}}.$$ Otherwise, if $n - \alpha p \le 0$ when you sum over $k$ you get
$$\int_{\{|x| < 1\}} |x|^{-\alpha p} \, dx = \infty.$$
Incidentally, when $n - \alpha p > 0$ you can take the limit as $t \to 1^-$ and apply the squeeze theorem to get the result $$\int_{\{|x| < 1\}} |x|^{-\alpha p} \, dx = \frac{\omega n}{n - \alpha p}$$ so that $$\|f_0\|_p = \left( \frac{\omega n}{n - \alpha p} \right)^{1/p}.$$
