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<syntaxhighlight lang="latex">\section{A conductance--connectivity--degree lower bound on the margin}

</syntaxhighlight>

and ending at its closing remark.

This version does not assume \hat x_C>0 anywhere. Instead, it defines x^\hat xx^ as the restricted minimizer on CCC, proves x^C≥0\hat x_C\ge 0x^C≥0, builds the global KKT certificate using the restricted KKT subgradient, and then derives an explicit parameterized lower bound on the outside margin on Cˉ\bar CCˉ in terms of ϕ(C)\phi(C)ϕ(C), μC\mu_CμC, and doutd_{\mathrm{out}}dout. It also includes a corollary giving an Ω(ρ)\Omega(\sqrt{\rho})Ω(ρ) regime.

<syntaxhighlight lang="latex">% ============================================================
\section{A conductance--connectivity--degree lower bound on the margin}
\label{sec:gamma_conductance_connectivity_degree}

We specialize to the $\ell_1$-regularized PageRank objective~(3):
\begin{equation}\label{eq:ppr_obj_gamma_param}
\min_{x\in\mathbb{R}^n}
F(x)
:=
\underbrace{\frac12 x^\top Qx-\alpha x^\top D^{-1/2}s}_{=:f(x)}
\;+\;
\underbrace{\rho\alpha\|D^{1/2}x\|_1}_{=:g(x)}.
\end{equation}
Here $G=(V,E)$ is undirected and unweighted, $A$ is the adjacency matrix, $D$ is the diagonal degree matrix,
$s=e_v$ is a seed at node $v$, and
\begin{equation}\label{eq:Q_def_gamma_param}
Q=\alpha I+\frac{1-\alpha}{2}L
=
\frac{1+\alpha}{2}I-\frac{1-\alpha}{2}\,D^{-1/2}AD^{-1/2}.
\end{equation}
The weighted $\ell_1$ penalty has coordinate weights
\begin{equation}\label{eq:lambda_def_gamma_param}
g(x)=\sum_{i=1}^n \lambda_i |x_i|,
\qquad
\lambda_i:=\rho\alpha\sqrt{d_i}.
\end{equation}
Since $Q\succeq \alpha I$, the function $f$ is $\alpha$-strongly convex and hence $F=f+g$ has a unique minimizer $x^\star$.

\paragraph{Cluster notation.}
Fix a set (candidate cluster) $C\subseteq V$ with $v\in C$, and let $\bar C:=V\setminus C$.
Define the volume and cut:
\[
\mathrm{vol}(C):=\sum_{i\in C} d_i,
\qquad
\mathrm{cut}(C,\bar C):=|\{(i,j)\in E:\; i\in C,\ j\in\bar C\}|.
\]
We use the (one-sided) conductance
\[
\phi(C):=\frac{\mathrm{cut}(C,\bar C)}{\mathrm{vol}(C)}.
\]
Let
\[
d_{\mathrm{in}}:=\min_{j\in C} d_j,
\qquad
d_{\mathrm{out}}:=\min_{i\in\bar C} d_i,
\qquad
d_v:=d_{v}.
\]
We write $N(i)$ for the neighbor set of node $i$.

\paragraph{Internal connectivity parameter.}
Let
\begin{equation}\label{eq:muC_def_gamma_param}
\mu_C:=\lambda_{\min}(Q_{CC}),
\end{equation}
the smallest eigenvalue of the principal submatrix $Q_{CC}$.

\subsection{Restricted minimizer on $C$, its nonnegativity, and a norm bound}

Define $b:=\alpha D^{-1/2}s$. Since $s=e_v$ and $v\in C$, the vector $b_C$ has exactly one nonzero entry and
\begin{equation}\label{eq:bC_norm_gamma_param}
\|b_C\|_2=\frac{\alpha}{\sqrt{d_v}}.
\end{equation}

\begin{definition}[Restricted minimizer on $C$]\label{def:xhat_restricted_gamma_param}
Let $\hat x$ denote the unique minimizer of \eqref{eq:ppr_obj_gamma_param} subject to $x_{\bar C}=0$:
\begin{equation}\label{eq:xhat_def_gamma_param}
\hat x \in \operatorname*{argmin}_{x\in\mathbb{R}^n:\; x_{\bar C}=0}\; F(x).
\end{equation}
Equivalently, $\hat x_{\bar C}=0$ and $\hat x_C$ uniquely minimizes
\[
\min_{x_C\in\mathbb{R}^{|C|}}
\left\{
\frac12 x_C^\top Q_{CC}x_C - b_C^\top x_C + \sum_{i\in C}\lambda_i|x_i|
\right\}.
\]
\end{definition}

\begin{lemma}[Absolute-value improvement and nonnegativity]\label{lem:abs_improvement_nonneg_gamma_param}
Assume $v\in C$ so that $b_C\ge 0$ and $b_{\bar C}=0$.
Then for every $x$ with $x_{\bar C}=0$, the componentwise absolute value $|x|$ satisfies
\[
F(|x|)\le F(x).
\]
Consequently, the restricted minimizer satisfies $\hat x_C\ge 0$ componentwise.
\end{lemma}

\begin{proof}
Fix any $x$ with $x_{\bar C}=0$ and set $y:=|x|$ (so $y_{\bar C}=0$).
Write
\[
F(x)=\frac12 x^\top Qx - b^\top x + \sum_{i=1}^n \lambda_i|x_i|.
\]

\medskip\noindent\textbf{Quadratic term.}
Since $Q=\frac{1+\alpha}{2}I-\frac{1-\alpha}{2}D^{-1/2}AD^{-1/2}$, we have $Q_{ij}\le 0$ for $i\neq j$.
Expand
\[
x^\top Qx = \sum_i Q_{ii}x_i^2 + 2\sum_{i<j}Q_{ij}x_i x_j.
\]
For each $i<j$, we have $x_i x_j\le |x_i||x_j|=y_i y_j$, and since $Q_{ij}\le 0$ this implies
$Q_{ij}x_i x_j \ge Q_{ij}y_i y_j$.
Hence $x^\top Qx \ge y^\top Qy$.

\medskip\noindent\textbf{Linear term.}
Since $b\ge 0$ and $y_i=|x_i|\ge x_i$ for all $i$, we have $-b^\top y \le -b^\top x$.

\medskip\noindent\textbf{$\ell_1$ term.}
$\sum_i \lambda_i|y_i|=\sum_i\lambda_i|x_i|$ by definition.

\medskip\noindent
Combining the three comparisons gives $F(y)\le F(x)$.
Now let $\hat x$ be the unique restricted minimizer. Then $F(|\hat x|)\le F(\hat x)$, so $|\hat x|$ is also a minimizer.
By uniqueness, $|\hat x|=\hat x$, hence $\hat x_C\ge 0$.
\end{proof}

\begin{lemma}[Restricted KKT on $C$]\label{lem:restricted_KKT_on_C_gamma_param}
There exists a vector $z_C\in\mathbb{R}^{|C|}$ such that
\begin{equation}\label{eq:restricted_KKT_on_C_gamma_param}
Q_{CC}\hat x_C - b_C + z_C = 0,
\end{equation}
and for every $i\in C$,
\begin{equation}\label{eq:zC_in_subdiff_gamma_param}
z_{C,i}\in \lambda_i\,\partial|\hat x_i| \subseteq [-\lambda_i,\lambda_i].
\end{equation}
In particular, $\|z_C\|_2 \le \|\lambda_C\|_2 = \rho\alpha\sqrt{\mathrm{vol}(C)}$.
\end{lemma}

\begin{proof}
The restricted objective over $x_C$ is strongly convex because $Q_{CC}\succ 0$ and the $\ell_1$ term is convex,
so $\hat x_C$ is unique. First-order optimality yields
\[
0\in Q_{CC}\hat x_C - b_C + \partial\!\Big(\sum_{i\in C}\lambda_i|\hat x_i|\Big),
\]
which is exactly the existence of $z_C$ satisfying \eqref{eq:restricted_KKT_on_C_gamma_param}--\eqref{eq:zC_in_subdiff_gamma_param}.
The norm bound follows from $|z_{C,i}|\le \lambda_i$ and
$\|\lambda_C\|_2^2=\sum_{i\in C}(\rho\alpha\sqrt{d_i})^2=(\rho\alpha)^2\mathrm{vol}(C)$.
\end{proof}

\begin{lemma}[Bounding $\|\hat x_C\|_2$ by internal connectivity]\label{lem:xC_norm_gamma_param}
Let $\mu_C=\lambda_{\min}(Q_{CC})$. Then
\begin{equation}\label{eq:xC_norm_gamma_param}
\|\hat x_C\|_2
\;\le\;
\frac{1}{\mu_C}\left(\frac{\alpha}{\sqrt{d_v}}+\rho\alpha\sqrt{\mathrm{vol}(C)}\right).
\end{equation}
\end{lemma}

\begin{proof}
From Lemma~\ref{lem:restricted_KKT_on_C_gamma_param}, $Q_{CC}\hat x_C=b_C-z_C$, so
\[
\|\hat x_C\|_2 \le \|Q_{CC}^{-1}\|_2\,\|b_C-z_C\|_2
\le \|Q_{CC}^{-1}\|_2\bigl(\|b_C\|_2+\|z_C\|_2\bigr).
\]
Since $\lambda_{\min}(Q_{CC})=\mu_C$, we have $\|Q_{CC}^{-1}\|_2=1/\mu_C$.
Also, $\|b_C\|_2=\alpha/\sqrt{d_v}$ by \eqref{eq:bC_norm_gamma_param}, and
$\|z_C\|_2\le \rho\alpha\sqrt{\mathrm{vol}(C)}$ by Lemma~\ref{lem:restricted_KKT_on_C_gamma_param}.
Substituting yields \eqref{eq:xC_norm_gamma_param}.
\end{proof}

\subsection{A pointwise outside-gradient bound from conductance and degrees}

\begin{lemma}[Outside gradient bound from conductance and degrees]\label{lem:outside_grad_bound_gamma_param}
Assume $v\in C$ so that $b_{\bar C}=0$. Then for each $i\in\bar C$,
\begin{equation}\label{eq:outside_grad_pointwise_gamma_param}
|\nabla f(\hat x)_i|
\le
\frac{1-\alpha}{2}\,\|\hat x_C\|_2\,
\sqrt{\frac{\mathrm{cut}(C,\bar C)}{d_i\,d_{\mathrm{in}}}}.
\end{equation}
Consequently, using $d_i\ge d_{\mathrm{out}}$ and $\mathrm{cut}(C,\bar C)=\phi(C)\mathrm{vol}(C)$,
\begin{equation}\label{eq:outside_grad_uniform_gamma_param}
\max_{i\in\bar C}|\nabla f(\hat x)_i|
\le
\frac{1-\alpha}{2}\,\|\hat x_C\|_2\,
\sqrt{\frac{\phi(C)\mathrm{vol}(C)}{d_{\mathrm{out}}\,d_{\mathrm{in}}}}.
\end{equation}
\end{lemma}

\begin{proof}
Fix $i\in\bar C$. Since $\hat x_{\bar C}=0$ and $b_{\bar C}=0$,
\[
\nabla f(\hat x)_i = (Q\hat x - b)_i = (Q_{\bar C C}\hat x_C)_i = \sum_{j\in C}Q_{ij}\hat x_j.
\]
From \eqref{eq:Q_def_gamma_param}, for $i\neq j$ we have
\[
Q_{ij} = -\frac{1-\alpha}{2}(D^{-1/2}AD^{-1/2})_{ij}
=
\begin{cases}
-\dfrac{1-\alpha}{2}\dfrac{1}{\sqrt{d_i d_j}}, & (i,j)\in E,\\[2mm]
0, & (i,j)\notin E.
\end{cases}
\]
Thus
\[
|\nabla f(\hat x)_i|
=
\left|\sum_{j\in C\cap N(i)} -\frac{1-\alpha}{2}\frac{\hat x_j}{\sqrt{d_i d_j}}\right|
\le
\frac{1-\alpha}{2}\cdot \frac{1}{\sqrt{d_i}}
\sum_{j\in C\cap N(i)}\frac{|\hat x_j|}{\sqrt{d_j}}.
\]
Apply Cauchy--Schwarz:
\[
\sum_{j\in C\cap N(i)}\frac{|\hat x_j|}{\sqrt{d_j}}
\le
\left(\sum_{j\in C\cap N(i)} \hat x_j^2\right)^{1/2}
\left(\sum_{j\in C\cap N(i)} \frac{1}{d_j}\right)^{1/2}
\le
\|\hat x_C\|_2\left(\sum_{j\in C\cap N(i)} \frac{1}{d_j}\right)^{1/2}.
\]
Since $d_j\ge d_{\mathrm{in}}$ for all $j\in C$,
\[
\sum_{j\in C\cap N(i)} \frac{1}{d_j}
\le
\frac{|C\cap N(i)|}{d_{\mathrm{in}}}.
\]
Finally, $|C\cap N(i)|$ is the number of cut edges incident to $i$, hence $|C\cap N(i)|\le \mathrm{cut}(C,\bar C)$,
which gives \eqref{eq:outside_grad_pointwise_gamma_param}. The uniform bound
\eqref{eq:outside_grad_uniform_gamma_param} follows by using $d_i\ge d_{\mathrm{out}}$
and $\mathrm{cut}(C,\bar C)=\phi(C)\mathrm{vol}(C)$.
\end{proof}

\subsection{A parameterized outside-margin lower bound and a $\sqrt{\rho}$ regime}

Define the cluster-dependent quantity
\begin{equation}\label{eq:DeltaC_def_gamma_param}
\Delta(C)
:=
\frac{1-\alpha}{2\mu_C}\,
\sqrt{\frac{\phi(C)\mathrm{vol}(C)}{d_{\mathrm{out}}\,d_{\mathrm{in}}}}\,
\left(\frac{\alpha}{\sqrt{d_v}}+\rho\alpha\sqrt{\mathrm{vol}(C)}\right).
\end{equation}

\begin{theorem}[Margin bound parameterized by conductance, internal connectivity, and outside degree]
\label{thm:gamma_parameterized_by_cluster}
Assume $v\in C$ and let $\hat x$ be the restricted minimizer \eqref{eq:xhat_def_gamma_param}.
Define the \emph{outside margin} on the complement
\[
\gamma_{\bar C}
:=
\min_{i\in \bar C}\Bigl(\lambda_i-|\nabla f(x^\star)_i|\Bigr).
\]
Then:
\begin{enumerate}
\item[(i)] If $\Delta(C)<\min_{i\in\bar C}\lambda_i=\rho\alpha\sqrt{d_{\mathrm{out}}}$, then $\hat x$ is the \emph{unique}
global minimizer of \eqref{eq:ppr_obj_gamma_param}. In particular, $x^\star=\hat x$ and $x^\star_{\bar C}=0$.
\item[(ii)] Whenever $x^\star=\hat x$, we have the explicit bound
\begin{equation}\label{eq:gamma_param_bound_gamma_param}
\gamma_{\bar C}
\;\ge\;
\rho\alpha\sqrt{d_{\mathrm{out}}}-\Delta(C).
\end{equation}
Equivalently, if $\Delta(C)\le (1-\eta)\rho\alpha\sqrt{d_{\mathrm{out}}}$ for some $\eta\in(0,1)$, then
\begin{equation}\label{eq:gamma_eta_bound_gamma_param}
\gamma_{\bar C} \ge \eta\,\rho\alpha\sqrt{d_{\mathrm{out}}}.
\end{equation}
\end{enumerate}
\end{theorem}

\begin{proof}
We prove (i) and (ii) in order.

\medskip\noindent\textbf{Step 1: KKT conditions.}
Since $f$ is differentiable and $g(x)=\sum_i \lambda_i|x_i|$, a point $x$ is optimal if and only if there exists
$z\in\partial g(x)$ such that $\nabla f(x)+z=0$.
Moreover, $F$ is strongly convex (because $Q\succeq \alpha I$), hence the minimizer is unique.

\medskip\noindent\textbf{Step 2: Build a global KKT certificate from the restricted minimizer.}
From Lemma~\ref{lem:restricted_KKT_on_C_gamma_param}, there exists $z_C$ such that
\[
Q_{CC}\hat x_C-b_C+z_C=0
\quad\text{and}\quad
z_{C,i}\in[-\lambda_i,\lambda_i]\;\; \forall i\in C.
\]
Define $z\in\mathbb{R}^n$ by
\[
z_i :=
\begin{cases}
(z_C)_i, & i\in C,\\[1mm]
-\nabla f(\hat x)_i, & i\in\bar C.
\end{cases}
\]
We claim $z\in\partial g(\hat x)$ if $|\nabla f(\hat x)_i|\le \lambda_i$ for all $i\in\bar C$.
Indeed:
\begin{itemize}
\item If $i\in C$, then $z_{C,i}\in \lambda_i\,\partial|\hat x_i|$ by \eqref{eq:zC_in_subdiff_gamma_param}, hence it is valid.
\item If $i\in\bar C$, then $\hat x_i=0$ and $z_i\in[-\lambda_i,\lambda_i]$ holds if
$|z_i|=|\nabla f(\hat x)_i|\le \lambda_i$.
\end{itemize}

Now check $\nabla f(\hat x)+z=0$.
On $C$, since $\hat x_{\bar C}=0$,
\[
\nabla f(\hat x)_C=(Q\hat x-b)_C = Q_{CC}\hat x_C - b_C,
\]
so $\nabla f(\hat x)_C+z_C=0$ by $Q_{CC}\hat x_C-b_C+z_C=0$.
On $\bar C$, $\nabla f(\hat x)_{\bar C}+z_{\bar C}=0$ by definition.
Thus $\nabla f(\hat x)+z=0$.

\medskip\noindent\textbf{Step 3: Enforce the outside inequalities via $\Delta(C)$.}
By Lemma~\ref{lem:outside_grad_bound_gamma_param} and Lemma~\ref{lem:xC_norm_gamma_param},
\[
\max_{i\in\bar C}|\nabla f(\hat x)_i|
\le
\Delta(C).
\]
Therefore, if $\Delta(C)<\rho\alpha\sqrt{d_{\mathrm{out}}}\le \lambda_i$ for all $i\in\bar C$,
then $|\nabla f(\hat x)_i|<\lambda_i$ on $\bar C$ and hence $z\in\partial g(\hat x)$.
Thus $0\in\partial F(\hat x)$, so $\hat x$ is a minimizer.
By uniqueness, $\hat x=x^\star$, proving (i).

\medskip\noindent\textbf{Step 4: Lower bound the outside margin.}
Assume $x^\star=\hat x$. Then for every $i\in\bar C$,
\[
\lambda_i-|\nabla f(x^\star)_i|
=\lambda_i-|\nabla f(\hat x)_i|
\ge \lambda_i-\Delta(C)
\ge \rho\alpha\sqrt{d_{\mathrm{out}}}-\Delta(C),
\]
where we used $\lambda_i\ge \rho\alpha\sqrt{d_{\mathrm{out}}}$.
Taking the minimum over $i\in\bar C$ yields \eqref{eq:gamma_param_bound_gamma_param}, and \eqref{eq:gamma_eta_bound_gamma_param} follows.
\end{proof}

\begin{corollary}[A concrete $\gamma_{\bar C}=\Omega(\sqrt{\rho})$ regime]\label{cor:gamma_Omega_sqrt_rho_regime}
Assume the hypotheses of Theorem~\ref{thm:gamma_parameterized_by_cluster}.
Suppose, in addition, that there are constants $V_0,d_0,\mu_0,c_0>0$ such that
\[
\mathrm{vol}(C)\le \frac{V_0}{\rho},\qquad
d_{\mathrm{in}}\ge d_0,\qquad
d_v\ge d_0,\qquad
\mu_C\ge \mu_0,\qquad
d_{\mathrm{out}}\ge \frac{c_0}{\alpha^2\rho}.
\]
If furthermore the conductance satisfies
\begin{equation}\label{eq:phi_rho_condition_gamma_param}
\phi(C)\le c_1\rho
\quad\text{for a sufficiently small constant $c_1$ (depending only on $V_0,d_0,\mu_0,c_0,\alpha$),}
\end{equation}
then there exists a constant $\eta\in(0,1)$ (depending only on the same parameters) such that
\[
\gamma_{\bar C} \ge \eta\,\sqrt{\rho}.
\]
\end{corollary}

\begin{proof}
First, from $d_{\mathrm{out}}\ge c_0/(\alpha^2\rho)$ we get
\[
\rho\alpha\sqrt{d_{\mathrm{out}}}\ge \rho\alpha\sqrt{\frac{c_0}{\alpha^2\rho}}=\sqrt{c_0}\,\sqrt{\rho}.
\]
Next, using the assumed bounds in \eqref{eq:DeltaC_def_gamma_param},
\begin{align*}
\Delta(C)
&\le
\frac{1-\alpha}{2\mu_0}\,
\sqrt{\frac{\phi(C)\cdot (V_0/\rho)}{(c_0/(\alpha^2\rho))\cdot d_0}}\,
\left(\frac{\alpha}{\sqrt{d_0}}+\rho\alpha\sqrt{V_0/\rho}\right)\\
&=
\frac{1-\alpha}{2\mu_0}\,
\alpha\,\sqrt{\frac{V_0}{c_0 d_0}}\,
\sqrt{\phi(C)}\,
\left(\frac{\alpha}{\sqrt{d_0}}+\alpha\sqrt{V_0}\sqrt{\rho}\right).
\end{align*}
Under \eqref{eq:phi_rho_condition_gamma_param}, $\sqrt{\phi(C)}\le \sqrt{c_1}\sqrt{\rho}$, hence
$\Delta(C)\le K\sqrt{c_1}\sqrt{\rho}$ for a constant $K$ depending only on $(V_0,d_0,\mu_0,c_0,\alpha)$.
Choosing $c_1$ sufficiently small ensures $\Delta(C)\le (1-\eta)\sqrt{c_0}\sqrt{\rho}$ for some $\eta\in(0,1)$, i.e.
\[
\Delta(C)\le (1-\eta)\rho\alpha\sqrt{d_{\mathrm{out}}}.
\]
Apply \eqref{eq:gamma_eta_bound_gamma_param} to conclude $\gamma_{\bar C}\ge \eta\rho\alpha\sqrt{d_{\mathrm{out}}}
\ge \eta\sqrt{c_0}\sqrt{\rho}$, and absorb $\sqrt{c_0}$ into $\eta$.
\end{proof}

\begin{remark}[Outside margin vs.\ the global margin $\gamma$]
Theorem~\ref{thm:gamma_parameterized_by_cluster} lower bounds the margin on $\bar C$.
The \emph{global} strict-complementarity margin $\gamma=\min_{i\in I^\star}(\lambda_i-|\nabla f(x^\star)_i|)$
could be smaller if some coordinates inside $C$ are also zero at $x^\star$.
To deduce the same lower bound for the global $\gamma$, it suffices (but is not implied by conductance alone)
to additionally have $x^\star_i>0$ for all $i\in C$, i.e.\ $I^\star=\bar C$.
\end{remark}
% ============================================================

</syntaxhighlight>

If you paste this in place of your current section, it should compile cleanly and removes the “mystery assumption” x^C>0\hat x_C>0x^C>0 entirely.