Lemma 12.6.39.1EF963CF5EA2 The end $v\to \infty $ of $I$ has finite length exactly when $r_0>1$, and the end $v\to -1/M$ exactly when $r_ M>1$. More precisely, at the end $v\to \infty $ put $f=a$, $\epsilon =1/v$ and $r=r_0$; at the end $v\to -1/M$ put $f=M-a$, $\epsilon =-1/v-M$ and $r=r_ M$. Let
\[ H_ f=\int _ B\frac{\sigma }{f+\epsilon },\quad J_ f=\int _ B\frac{\sigma }{(f+\epsilon )^2},\quad K_ f=\int _ B\frac{\sigma }{(f+\epsilon )^3}. \]
Then $\epsilon \downarrow 0$ at the end, and
DB02B7FF941A
12.6.40
\begin{equation} \rho (v)\, \mathrm{d}v^2=G_ f(\epsilon )\, \mathrm{d}\epsilon ^2, \quad G_ f=\frac2n\left[\frac{K_ f}{H_ f} -\left(\frac{J_ f}{H_ f}\right)^2\right]. \end{equation}
Moreover, $G_ f\asymp \epsilon ^{-2}$ for $r<1$ and $G_ f\asymp \epsilon ^{-2}/\log (1/\epsilon )$ for $r=1$, while for $r>1$ the function $\sqrt{G_ f}$ is integrable near $\epsilon =0$.
Proof.
At the end $v\to \infty $, where $\epsilon =1/v$, we have
\[ \ell _ v=\epsilon -\frac{\epsilon ^2}{a+\epsilon },\qquad \mathrm{d}v=-\epsilon ^{-2}\mathrm{d}\epsilon . \]
At the end $v\to -1/M$, where $\epsilon =-1/v-M$, we obtain
\[ \ell _ v=\frac{(M+\epsilon )^2}{M-a+\epsilon }-(M+\epsilon ), \qquad \mathrm{d}v=(M+\epsilon )^{-2}\mathrm{d}\epsilon . \]
In both cases $\ell _ v$ is an affine function of $1/(f+\epsilon )$ with coefficients depending only on $\epsilon $, and (12.6.40) expresses $\rho (v)\, \mathrm{d}v^2$ through the variance of $1/(f+\epsilon )$ under the probability $\sigma /[H_ f(f+\epsilon )]$. This probability is $\nu _ v$, the marginal in (12.6.35), because $L$ is a positive multiple of $f+\epsilon $ and positive scaling of $L$ does not change that marginal.
Integration over tubes around the components $Z_ i$, splitting the normal radius at $\epsilon ^{1/k_ i}$, gives
\[ H_ f\asymp \begin{cases} \epsilon ^{r-1}& r<1,\\ \log (1/\epsilon )& r=1,\\ 1& r>1, \end{cases} \qquad K_ f\asymp \begin{cases} \epsilon ^{r-3}& r<3,\\ \log (1/\epsilon )& r=3,\\ 1& r>3. \end{cases} \]
The analogous threshold for $J_ f$ is two. For a finite sum of tube integrals, the smallest ratio governs.
If $r<1$, choose a component attaining $r$, and two normal annuli at fixed multiples of $\epsilon ^{1/k_ i}$, so far apart that the two-sided bounds separate their ranges of $f/\epsilon $. Each annulus has probability bounded below, since its weighted mass and $H_ f$ both have order $\epsilon ^{r-1}$, and the values of $1/(f+\epsilon )$ on the two annuli differ by at least $c\epsilon ^{-1}$. The identity
\[ \operatorname {Var}_\nu T =\frac12\iint (T(x)-T(y))^2\, \mathrm{d}\nu (x)\mathrm{d}\nu (y) \]
applied to $T=1/(f+\epsilon )$ gives the lower bound, and $T\leq \epsilon ^{-1}$ gives the upper bound.
If $r=1$, the near annulus has probability at least $c/\log (1/\epsilon )$. In the same tube, the normal radii between $\epsilon ^{1/(2k_ i)}$ and a fixed small radius carry probability bounded below, by the logarithmic integral. There $1/(f+\epsilon )\leq C\epsilon ^{-1/2}$, while $1/(f+\epsilon )\geq c\epsilon ^{-1}$ on the near annulus. The variance identity and the bound $G_ f\leq 2K_ f/(nH_ f)$ prove the claimed logarithmic estimate. In both cases $r<1$ and $r=1$, the integral of $\sqrt{G_ f}$ diverges at $\epsilon =0$.
For $r>1$, $H_ f$ tends to a finite positive integral and $G_ f\leq C K_ f$. For $1<r<3$, $\sqrt{G_ f}$ is therefore bounded by $C\epsilon ^{(r-3)/2}$, whose exponent exceeds $-1$. At $r=3$ the bound is $C\sqrt{\log (1/\epsilon )}$, and for $r>3$ it is a constant. All these bounds are integrable.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.