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Define $f: [0, \infty[ \to \bar{\mathbb R}$ measurable

Show that:

$f \circ | \cdot |: \mathbb R^d \to \bar{\mathbb R}$ as a lebesgue integrable function $\iff$ $g: \mathbb R_{\geq 0}\to \mathbb R, g(r):=r^{d-1}f(r)$ is lebesgue integrable

Ideas:

"$\Rightarrow$" $g$ is simply $f$ scaled by a constant $r^{d-1}$ and therefore measurable and

$\int_{[0,\infty[}g(r)dr=\int_{[0,\infty[}r^{d-1}f(r)dr$

I would like to say that find a constant $c \in \mathbb R$ so that $|r^{d-1}|\leq c$ but obviously $r \in [0,\infty[$ so this approach would not make sense. The only other solution that comes to mind would be substitution, so

set $|x| = r \Rightarrow dx =dr$ (Is this even correct?) but here I do not get any further.

"$\Leftarrow$"

$f \circ |\cdot|$ is by definition measurable

All I notice is that on $r \in [1,\infty[$: $f(r) \leq g(r)$ and then using the monotonicity of integrals I could use $\int_{[1,\infty[}f(r)dr \leq \int_{[1,\infty[}g(r)dr < \infty$ but this still does not help me on $]0,1[$

Any help would be greatly appreciated.

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  • $\begingroup$ No, $f \circ |\cdot|: \mathbb R^d \to \bar{\mathbb R}$, while $f: [0, \infty[ \to \bar{\mathbb R}$ $\endgroup$ – SABOY Jan 4 '19 at 20:02
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    $\begingroup$ Do you know the "polar coordinates" formula? The case $d=2$ should be familiar. $\endgroup$ – John Dawkins Jan 4 '19 at 23:05
  • $\begingroup$ See Rudin's RCA for polar coordinates in $\mathbb R^{n}$. $\endgroup$ – Kavi Rama Murthy Jan 5 '19 at 0:03
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As you have already seen measurability of $f$ and $g$ is not a problem. It is enough to show that for $f \geq 0$ we have $\newcommand{\RR}{\mathbb{R}}$

$$I := \int_{\mathbb{R}^n} f(|x|) \,d x < \infty \iff J := \int_\mathbb{R_{\geq0}} r^{d-1} f(r) \, d r <\infty$$

This is integration using $d$-dimensional spherical coordinates. They provide a diffeomorphism $\Psi$ from $\mathbb{R}_{>0} \times (0, \pi)^{d-2} \times (0, 2\pi)$ to $\mathbb{R}^n \setminus N$ where $N$ is some set of measure zero ($N$ is something like $\mathbb{R}_{>0} \times \{0\} \times \mathbb{R}^{d-2}$, but it depends on the details of the parametrization and doesn't really matter). The special thing about spherical coordinates is that $|\Psi(r, \phi_1, \dots, \phi_{d-1})| = r$

The Jacobi determinant of this map is $\det(\mathrm{D} \Psi(x)) = r^{d-1} \sin^{n-2} \phi_1 \sin^{n-3} \phi_2 \dots \sin \phi_{d-1}$. Therefore, by the change of variables formula

$$ \begin{align} I &= \int_{\RR_{>0}} \int_{(0, \pi)^{d-2}} \int_{(0, 2 \pi)} f(|\Psi(r, \phi_1, \dots, \phi_{d-1})|)\det(\mathrm{D} \Psi(x)) \,d \phi_{d-1} \,d \phi_{d-2} \dots \,d \phi_1 \,d r \\ &= \int_{\RR_{>0}} \int_{(0, \pi)^{d-2}} \int_{(0, 2 \pi)} f(r) r^{d-1} \sin^{n-2} \phi_1 \sin^{n-3} \phi_2 \dots \sin \phi_{d-1} \,d \phi_{d-1} \,d \phi_{d-2} \dots \,d \phi_1 \,d r \\ &= \int_{\RR_{>0}} f(r) r^{d-1} \, dr \cdot \int_{(0, \pi)^{d-2}} \int_{(0, 2 \pi)} \sin^{n-2} \phi_1 \sin^{n-3} \phi_2 \dots \sin \phi_{d-1} \,d \phi_{d-1} \,d \phi_{d-2} \dots \,d \phi_1 \\ &= \int_{\RR_{>0}} f(r) r^{d-1} \, dr \cdot K = J \cdot K \end{align}$$ where $K$ is the second integral in the line before the last. One clearly sees that the absolute value of the integrand in $K$ is bounded by $1$. Therefore $K \leq 2 \pi^{d-1}$. So $I$ and $J$ are the same up to a finite constant and therefore $I < \infty \iff J < \infty$.

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