\section{Z is nowhere-dense}
\label{Sec_LimitAnalysis}

Since $\int_M |\Psi|^2 = 1$, we know that $Z$ cannot be the entire space.
To obtain more precise information on $Z$ it turns out to be helpful to apply the ideas from \autoref{Sec_Frequency} to the limit $(A,\Psi)$.
Fix $x \in M$ and define functions $H,h \co [0,r_0] \to [0,\infty)$ by 
\begin{align*}
  H(r) &:= \int_{B_r(x)} |\nabla_{A\otimes B} \Psi|^2 \qand \\
  h(r) &:= \int_{\del B_r(x)} |\Psi|^2.
\end{align*}
Here we extend $|\nabla_{A\otimes B} \Psi|$ by defining it to be zero on $Z$.
If $h(r) > 0$, define
\begin{equation*}
  \scN(r):=\frac{rH(r)}{h(r)}.
\end{equation*}

\begin{prop}
  \label{Prop_UniformConvergenceOfhiHi}
  Denote by $h_i$, $H_i$ the $(A_i,\Psi_i,\alpha_i)$ version of $h$ and $H$ defined in \autoref{Def_Frequency}.
  The sequences of functions $h_i$ and $H_i$ converge uniformly to $h$ and $H$, respectively.
  In particular, $\scN_i(r) \to \scN(r)$ whenever $h(r) > 0$.
\end{prop}

Let us first explain how this implies the following.

\begin{prop}
  \label{Prop_ZIsNowhereDense}
  $Z$ is nowhere-dense.
\end{prop}

\begin{proof}
  Choose $R \geq 0$ as large as possible, but so that $B_R(x) \subset Z$.
  We know that $R$ is finite, because $Z$ is compact.
  By replacing $x$ with a point close to the boundary of $B_R(x)$ we can assume that $R \ll 1$.
  By construction of $R$ there is an $\epsilon \ll 1$ such that $h(R+\epsilon) > 0$.
  In particular, $\scN(R+\epsilon)$ is defined.
  It follows from \autoref{Prop_NControlsGrowthOfh} and \autoref{Prop_UniformConvergenceOfhiHi} that $R=0$.
\end{proof}

\begin{proof}[Proof of \autoref{Prop_UniformConvergenceOfhiHi}]
  That $h_i$ converges uniformly to $h$ is a direct consequence of the $C^{0,\gamma}$ convergence of $|\Psi_i|$.
  The proof of the corresponding statement for $H_i$ has three steps.

  \setcounter{step}{0}
  \begin{step}
    \label{Step_UniformConvergenceofHepsiloni}
    For $\epsilon \in (0,1/2]$ set $Z_\epsilon := |\Psi|^{-1}([0,\epsilon])$.
    The sequence of functions
    \begin{equation*}
      H_{\epsilon,i}(r) := \int_{B_r(x) \setminus Z_\epsilon} |\nabla_{A_i\otimes B} \Psi_i|^2 + \tan(\alpha_i)^{-2} |\mu(\Psi_i)|^2
    \end{equation*}
    converges uniformly to
    \begin{equation*}
      H_{\epsilon}(r) := \int_{B_r \setminus Z_\epsilon} |\nabla_{A\otimes B} \Psi|^2.
    \end{equation*}
  \end{step}

  This follows from the facts that $\tan(\alpha_i)^{-1}\mu(\Psi_i) = \tan(\alpha_i)F_{A_i}$ converges to zero in $L^2(M \setminus Z_\epsilon)$ and $\nabla_{A_i\otimes B}\Psi_i$ converges to $\nabla_{A\otimes B} \Psi$ in $L^2(M \setminus Z_\epsilon)$, see \autoref{Prop_Convergence}.

  \begin{step}
    \label{Step_HiMinusHepsiloni}
    There exists a $\lambda > 0$ such that
    \begin{equation*}
    \int_{Z_\epsilon} |\nabla_{A_i\otimes B} \Psi_i|^2 + \tan(\alpha_i)^{-2}|\mu(\Psi_i)|^2 = O(\epsilon^\lambda).
  \end{equation*}
  \end{step}

  Fix a cut-off function $\chi\co \R \to [0,1]$ with $\chi(t)=1$ for $t\leq 1$ and $\chi(t)=0$ for $t\geq 2$.
  Applying \autoref{Prop_IntegrationByParts} with $f=\chi(\epsilon^{-1}|\Psi_i|)$ and $U=M$, integrating the resulting term with $\Delta|\Psi|$ by parts once and using Kato's inequality yields
  \begin{align*}
    \int_{Z_\epsilon} |\nabla_{A_i\otimes B}\Psi_i|^2 + \tan(\alpha_i)^{-2}|\mu(\Psi_i)|^2 \leq c\epsilon^2 + c \int_{Z_{2\epsilon}\setminus Z_{\epsilon}} |\nabla_{A_i\otimes B}\Psi_i|^2.
  \end{align*}
  Denoting
  \begin{equation*}
    f(\epsilon) := \int_{Z_\epsilon} |\nabla_{A_i\otimes B}\Psi_i|^2 + \tan(\alpha_i)^{-2}|\mu(\Psi_i)|^2
  \end{equation*}
  this can be written as
  \begin{equation*}
    f(\epsilon) \leq \sigma (\epsilon^2 + f(2\epsilon))
  \end{equation*}
  with $\sigma := c/(1+c)$.
  Since $f$ is bounded above and we can assume that $\sigma \geq 1/2$,
  \begin{align*}
    f(\epsilon)
    &\leq \sigma\epsilon^2 \sum_{i=0}^{k-1} (4\sigma)^i + \sigma^kf(2^k\epsilon) \\
    &\leq \epsilon^2 \sigma\(\frac{(4\sigma)^{k-1}-1}{4\sigma-1}\) + c\sigma^k \\
    &\lesssim \epsilon^2(4\sigma)^k + \sigma^k.
  \end{align*}
  With $k := \floor{-\log \epsilon/\log 2}$ this gives
  \begin{align*}
    f(\epsilon) \lesssim \epsilon^{2-\log(4\sigma)/\log 2} + \epsilon^{-\log\sigma/\log 2} \lesssim \epsilon^\lambda
  \end{align*}
  for some $\lambda>0$ depending on $\sigma$ only, since $\log(4\sigma)/\log 2 < 2$.

  \begin{step}
  The sequence of functions $H_i$ converges uniformly to $H$. 
  \end{step}

  Both $|H_{\epsilon}(r)-H(r)|$ and $|H_{\epsilon,i}(r)-H_i(r)|$ converge uniformly to zero as $\epsilon$ goes to zero, the former by monotone convergence and the latter by \autoref{Step_HiMinusHepsiloni};
  hence, the desired convergence follows immediately from \autoref{Step_UniformConvergenceofHepsiloni}.
\end{proof}
{"config": "arxiv", "file": "1406.5683/ZnowhereDense.tex"}