Tips on finding the correct upper bound for $\frac{1}{\vert x\vert(1+\vert x\vert)^d}$

Question

Let $f: \mathbb R^{d} \to \mathbb R$ so that $x \mapsto \begin{cases} \frac{1}{\vert x\vert(1+\vert x \vert)^d }, x \neq 0\\ 0, x = 0 \end{cases}$

Determine for which $p\in [1,\infty], f \in\mathbb L^{p}(\mathbb R^{d})$

My idea:

Since since all of the $x$ are in absolute value and thus positive, I can reduce this problem to the function: $f: [0,\infty] \to \mathbb R$

and note that

$\frac{1}{ x^{p}(1+ x )^{dp} }1_{(1,\infty)}(x)\leq\frac{1}{x^{p}}1_{(1,\infty)}(x)$ which is in $L^{p}$ for $ p > 1$

Any idea on the region $(0,1)$?

Thanks for the help

Answer 1

You are almost right. Remember, though, that when you pass to spherical coordinates (this is what you do in your post), the underlying Lebesgue measure of $\mathbb R^n$ changes too. More clearly, for $p \ne \infty$

$$\int _{\mathbb R^d} \frac 1 {|x|^p (1 + |x|)^{pd}} \ \mathrm d x = \int _0 ^\infty \int _{S^{d-1}} \frac 1 {r^p (1+r)^{pd}} \ r^{d-1} \ \mathbb d s \ \mathrm d r = A \int _0 ^\infty \frac 1 {r^{p-d+1} (1+r)^{pd}} \ \mathrm d r$$

where $A = \int _{S^{d-1}} 1 \ \mathrm d s$ is the measure of the unit sphere $S^{d-1}$ computed with the intrinsic measure $\mathrm d s$.

To check the finiteness of $\int _0 ^\infty$ split it in two:

• on $(0,1]$ the factor $\dfrac 1 {(1+r)^{pd}}$ does not matter, because it is continuous and bounded, therefore integrable, so the only one that matters is $\dfrac 1 {r^{p-d+1}}$, and it is easy to see that this one is integrable if and only if $p-d+1<1$, i.e. $p<d$;

• on $(1, \infty)$ the factor $\dfrac 1 {(1+r)^{pd}}$ behaves like $\dfrac 1 {r^{pd}}$ (because their quotient has limit $1$ when $r \to \infty$), so the limit comparison test for improper integrals tells us that your integral has the same behaviour as

  $$\int _1 ^\infty \frac 1 {r^{p-d+1 + pd}} \ \mathrm d r$$

  (i.e. both are simultaneously either divergent, or convergent), which is easily seen to be convergent if and only if $p-d+1 + pd > 1$, i.e. $p > \dfrac d {d+1}$, which is trivially true since it is given that $p>1 > \dfrac d {d+1}$.

Since $p$ must satisfy both conditions, it follows that $p < d$.

The case $p = \infty$ must be treated separately, because the above argument does not work then. In this case you want to know whether $\dfrac 1 {|x| (1+|x|)^d}$ is bounded. The factor $\dfrac 1 {(1+|x|)^d}$ clearly is, but the factor $\dfrac 1 {|x|}$ is not because of its unboundedness towards $0$.

We conclude that the only $p$ for which $f \in L^p (\mathbb R^d)$ are $p<d$.

Answer 2

Hint: For every $d$ we have $m_d(B(0,r))= r^dm_d(B(0,1))$ for all $r>0.$ It follows that if $0<a<b,$ then $$m_d(\{a<|x|<b\}) = (b^d-a^d)m_d(B(0,1)).$$ Consider integrating over the sets

$$E_n = \{2^n<|x|<2^{n+1}\},\,n\in \mathbb Z.$$