Proposition 5.16.461D6716B27F1

Proof. Note that (5.16.5) is trivial when $n=0$, so we may assume that $n>0$ in the proof.

It follows from proposition 5.13.11 that the limit in (5.16.5) exists and is finite. We shall denote the limit by $\nu '(\varphi ,0)$ for the time being.

We first observe that by proposition 5.13.11, for any fixed $r_1\in (0,1)$ there is a constant $C_{r_1}$ such that

when $x\in B_ n(0,r_1)\setminus \{ 0\} $. In particular, $\nu (\varphi ,0)\geq \nu '(\varphi ,0)$.

In order to argue the reverse inequality, we may assume that $\nu (\varphi ,0)>0$.

Next observe that by (5.16.2), for each small enough $\epsilon >0$, we can find $r_0\in (0,1)$ and $C>0$ so that for all $x\in B_ n(0,r_0)\setminus \{ 0\} $, we have

It follows that $\nu '(\varphi ,0)\geq \nu (\varphi ,0)-\epsilon $. Letting $\epsilon \to 0+$, we conclude that $\nu '(\varphi ,0)\geq \nu (\varphi ,0)$, which combined with the previous inequality gives (5.16.5). $\square$