Proposition 12.13.4.6E8682D7FD89 Suppose that $Z=\{ a=0\} $ is a nonempty smooth compact real submanifold of positive codimension $d$, that $a\asymp \operatorname {dist}(\cdot ,Z)^ k$ in a fixed tube for a real number $k>0$, and that $a$ is bounded below by a positive constant outside that tube. Put $r=d/k$. Then, as $\epsilon \downarrow 0$,
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12.13.5
\begin{equation} \mathcal C(t-\epsilon )\asymp \begin{cases} \epsilon ^{(1-r)/n-1},& r<1,\\ \epsilon ^{-1}(\log (1/\epsilon ))^{-(n+1)/n},& r=1,\\ \epsilon ^{r-2},& 1<r<2,\\ \log (1/\epsilon ),& r=2,\\ 1,& r>2. \end{cases} \end{equation}
Each rate is integrable, and the endpoint volume is positive exactly when $d>k$. For $r>2$, the cost tends to
\[ n^{-(n-1)/n}W^{1/n}H(0)^{-(n+1)/n}J(0)>0, \qquad J(0)=\int _ B\frac{\sigma }{a^2}. \]
For finitely many disjoint zero components satisfying these hypotheses with codimensions $d_ i$ and orders $k_ i$, the same assertions hold with $r=\min _ i d_ i/k_ i$.
Proof.
The positive density and the tubular Jacobians are bounded above and below by positive constants. For $p=1,2$, radial integration therefore gives
\[ \int _ B\frac{\sigma }{(a+\epsilon )^ p} \asymp 1+\int _0^ R\frac{u^{d-1}}{(u^ k+\epsilon )^ p}\, \mathrm{d}u =1+\frac{\epsilon ^{r-p}}k \int _0^{R^ k/\epsilon }\frac{v^{r-1}}{(1+v)^ p}\, \mathrm{d}v, \]
with the equality applying to the radial expression. For $r<p$ the last integral tends to a finite positive constant; for $r=p$ it grows logarithmically; for $r>p$ the unregularized moment is finite, and monotone convergence gives its positive limit. Applying this to $H$ and $J$ in (12.13.2) proves the five rates and the criterion $H(0)<\infty $ exactly when $r>1$.
The power $(1-r)/n-1$ exceeds $-1$ for $r<1$, as does $r-2$ for $1<r<2$. At $r=1$, substitution of $\log (1/\epsilon )$ gives an integrable tail because $(n+1)/n>1$. The logarithmic and bounded rates are also integrable. When $r>2$, both inverse moments have finite positive limits, which gives the displayed limit. With finitely many disjoint components, the moments are finite sums of positive tube contributions and a bounded complement term; their smallest ratio determines the divergence, without cancellation.
$\square$
Pome
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