I'm trying to understand a proof in Michel Willem, Functional Analysis -- Fundamentals and Applications, Birkhäuser. The book defines:

$$\int_{\Bbb S^{N-1}}f(\sigma)\,d\sigma=N\int_{B_N}f\left(\frac{x}{|x|}\right)dx.$$

And then goes on to proving:

$\textbf{Lemma 2.4.7.}$ Let $u\in{\cal K}(\Bbb R^N)$. Then

(a) for every $r>0$, the function $\sigma\mapsto u(r\sigma)$ belongs to $C(\Bbb S^{N-1})$;

(b) $\displaystyle\frac{d}{dr}\int_{|x|<r}u(x)\, dx=r^{N-1}\int_{\Bbb S^{N-1}} u(r\sigma)\,d\sigma;$

(c) $\displaystyle \int_{\Bbb R^N}u(x)\, dx=\int_0^\infty r^{N-1} d{r}\,\int_{\Bbb S^{N-1}}u(r\sigma)\,d\sigma.$

Proof. (a) The restriction of a continuous function is a continuous function.

(b) Let $w(r)=\int_{|x|<r}u(x)\, dx$ and $v(r)=\int_{\Bbb S^{N-1}}u(r\sigma)\, d\sigma, r>0$. By definition, we have $$v(r)=N\int_{B_N}u\left(\frac{r}{|x|}x\right)dx.$$ Choose $r>0$ and $\varepsilon>0$. By definition of the modulus of continuity, we have > \begin{align} \left|w(r+\varepsilon)-w(r)-\int_{r<|x|<r+\varepsilon}u(rx/|x|)\, dx\right|={}&\left|\displaystyle\int_{r<|x|<r+\varepsilon}u(x)-u(rx/|x|)\, dx\right| \\ &{}\leq \omega_u(\varepsilon)V_N[(r+\varepsilon)^N-r^N]. \end{align}

The preceding theorem implies that

$$\int_{r<|x|<r+\varepsilon}u(rx/|x|)\,dx=\int_{|x|<r+\varepsilon}u(rx/|x|)\, dx-\int_{|x|<r}u(rx/|x|)\, dx=\frac{(r+\varepsilon)^N-r^N}{N}v(r).$$

Hence we find that:


so that:


The right derivative of $w$ is equal to $r^{N-1}v$. Similarly, the left derivative of $w$ is equal to $r^{N-1}v$.

(c) The fundamental theorem of calculus implies that for $0<a<b$,


Taking the limit as $a\to0$ and $b\to+\infty$, we obtain (c).

The first inequality chain is clear: the $=$ is just rewriting the difference and joining two integrals, the $\leq$ is using that $x$ and $\frac{r}{|x|}x$ have distance less than $\epsilon$ since they are on the same line through the origin and $|x|$ is between $r$ and $r+\epsilon$ whereas $r\frac{x}{|x|}$ has norm $r$, hence the modulus of continuity, which is the sup of $|u(y)-u(x)|$ when $|y-x|<\epsilon$, then what is left is the difference of the volumes of the spheres, and the volumes are $V_Nr^N$, where $V_N$ is that of the unit ball, since the volume scales with the cube of the radius.

The second chain is mysterious. The first $=$ is OK, but what about the second? I tried reading "the preceding theorem", but I only found things about change of variables which I do not see how to apply here. What can I do to get that equality?

Then the result of equality 2 is substituted into inequality 1, implying that absolute value tends to 0 as $\epsilon\to0$. Then we divide by $\epsilon$, and the RHS will tend to 0 since the modulus of continuity does, whereas the square bracket divided by $\epsilon$ will tend to the derivative of $r^N$ in $r$, which is bounded. So the LHS divided by $\epsilon$ also tends to 0, and seen as that fraction is $\epsilon$ the derivative $r^{N-1}v(r)$ plus higher order terms, the LHS has the same limit as what is given below, hence the right derivative is the right value, and the left derivative will be proven similarly, hence (b) is proven.

At this point, (c) is easy.

So how do I get that equality 2?


The theorem just before this is:

$\textbf{Theorem 2.4.5.}$ Let $u\in{\cal L}^1(\omega)$. Then $u(f)|J_f|\in{\cal L}^1(\Omega)$, and $(\ast)$ is valid.

Proof. By assumption, there exist $v,w\in{\cal L}^+(\omega)$ such that $u=v-w$ almost everywhere on $\omega$. It follows that $$u(f)|J_f|=v(f)|J_f|-w(f)|J_f|$$ almost everywhere on $\Omega$. It is easy to conclude the proof using Lemma 2.4.3. $\tag*{$\square$}$

$(\ast)$ is the formula for the change of variables in an integral:

$\quad$ We assume that $f:\Omega\to\omega$ is a diffeomorphism. The next theorem is proved in Sect. 9.1.

$\textbf{Theorem 2.4.2.}$ Let $u\in{\cal K}(\omega)$. Then $u(f)|J_f|\in{\cal K}(\Omega)$ and $$\int_\Omega u(f(x))|J_f(x)|dx=\int_\omega u(y)du. \tag{$\ast$}$$

  • $\begingroup$ It might help if you also provide a copy of the theorem cited, but I'm under the impression that it is just a rescaling of the integrand(s), in order to integrate two times the same function on the same domain so you can tear out the integral which defines $\nu$ (or is it $v$?) out of the resulting expression $\endgroup$
    – Thomas
    Dec 23, 2015 at 17:25
  • $\begingroup$ @Thomas I believe it's a $v$. I am not sure what theorem it is citing, but I will add an image of the theorem just before the "polar coordinates" section. $\endgroup$
    – MickG
    Dec 23, 2015 at 17:26
  • $\begingroup$ is $\nu$ defined as integral over the ball of radius $1$ or the ball of radius $r$? $\endgroup$
    – Thomas
    Dec 23, 2015 at 17:43
  • $\begingroup$ $v$ is defined at the beginning of the proof, and the integral it is defined as is defined in the first picture. If we take $N$ to mean the $N$-dimensional ball, I guess the radius is assumed to be 1 because it is not written explicitly so it must be 1 :). Or the ball could be of radius $N$, but taking $N$ for the dimension seems more sensible :). $\endgroup$
    – MickG
    Dec 23, 2015 at 18:10

1 Answer 1


I think the radius of the ball in the definition of $\nu$ is $=1$. If this is true note that, if $$\nu(r)=N\int_{B(r)} u(r\frac{x}{|x|})\,dx $$ and $f(x) = sx$ from the theorem you get $$\nu(r) = N \int_{B(s^{-1}r)}u(r\frac{y}{|y|})s^n\, dy$$ Now choose $s=r+\varepsilon$ in the first integral and $s=r$ in the second (thereby transforming both integrals to the unit ball, at the cost of a factor).


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