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For what values of $1\le p \le \infty$ does $f(x,y)=\frac{1}{1+|x|+|y|}$ with $(x,y) \in \mathbb{R}^2$ belong to $L^p(\mathbb{R}^2)$?

Using Wolfram Alpha I've found that the answer should be $p > 2$, but I don't know how to begin proving it. I've thought about doing a change of variable, after considering that $\displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left( \frac{1}{1+|x|+|y|} \right) ^pdxdy=4\int_0^{\infty}\int_0^{\infty}\left( \frac{1}{1+x+y} \right) ^pdxdy$

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3 Answers

up vote 5 down vote accepted

Because of the $1$ in the denominator we only have to bother about the complement of the unit disk. As $r:=\sqrt{x^2+y^2}\leq|x|+|y|\leq 2r$ we have $$r\leq 1+|x|+|y|\leq 3r\qquad(r\geq 1)\ .$$ It follows that the integral $$\int_{{\mathbb R}^2}{1\over (1+|x|+|y|)^p}\ {\rm d}(x,y)$$ is finite iff the integral $$2\pi \int_1^\infty {r\over r\mathstrut^p}\ dr$$ is finite, and this is the case when $p>2$.

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For a conceptual proof, (1.) do not focus on the exact values of the integrals involved and (2.) turn to polar coordinates.

All norms on $\mathbb R^2$ are equivalent and, if $(x,y)=(r\cos\theta,r\sin\theta)$, then $r$ is the $\ell^2$ norm of $(x,y)$ while $|x|+|y|$ is its $\ell^1$ norm hence $ar\leqslant|x|+|y|\leqslant br$ for some absolute constants $a$ and $b$ (and it happens that $a=1$ and $b=\sqrt2$ but these values are anecdotal here). For every positive $c$, the $L^p(\mathbb R^2)$ norm of $(x,y)\mapsto1/(1+cr)$ is $$ \iint\frac{r\mathrm dr\mathrm d\theta}{(1+cr)^p}=2\pi\int_0^{+\infty}\frac{r\mathrm dr}{(1+cr)^p}. $$ The function in the last integral is locally integrable and equivalent to a multiple of $r^{1-p}$ at infinity, hence it is integrable if and only if $1-p\lt-1$, that is, $p\gt2$. Since this does not depend on $c$, the same holds for the original function of $(x,y)$.

Likewise, in $\mathbb R^d$, the function $(x_1,\ldots,x_d)\mapsto1/(1+|x_1|+\cdots+|x_d|)$ is in $L^p(\mathbb R^d)$ if and only if $p\gt d$.

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You started promising, you just need to integrate. To start, for $y>0$ and $p>1$, $$ \int_0^\infty \frac{dx}{(1+x+y)^p} = -\frac{1}{p-1} \left[\frac{1}{(1+x+y)^{p-1}}\right]_{x=0}^\infty = \frac{1}{(p-1)(1+y)^{p-1}}. $$ Then $$ \int_0^\infty \frac{dy}{(p-1)(1+y)^{p-1}} $$ converges iff $p>2$. (And you can even get the exact value of the integral.)

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