path: root/approximation.tex

Even though \EDP{} is NP-hard, designing a mechanism for this problem will
involve being able to find an approximation $\tilde{L}^*(c)$ of $OPT$
monotonous with respect to coordinate-wise changes of the cost: if $c$ and $c'$
are two cost vectors such that $c'=(c_i', c_{-i})$ with $c_i' \leq c_i$, then
we want $\tilde{L}(c')\geq \tilde{L}(c)$. Furthermore, we seek an approximation
that can be computed in polynomial time.

This approximation will be obtained by introducing a concave optimization
problem with a constant approximation ratio to \EDP{}
(Proposition~\ref{prop:relaxation}).  Using Newton's method, it is then
possible to solve this concave optimization problem to an arbitrary precision.
However, this approximation breaks the monotonicity of the approximation.
Finding a monotone approximate solution to the concave problem will be the
object of (Section~\ref{sec:monotonicity}).

\subsection{A concave relaxation of \EDP}\label{sec:concave}

Let us introduce a new function $L$:
\begin{equation}\label{eq:our-relaxation}
\forall\,\lambda\in[0,1]^n,\quad L(\lambda) \defeq
\log\det\left(I_d + \sum_{i\in\mathcal{N}} \lambda_i x_i\T{x_i}\right),
\end{equation}

This function is a relaxation of the value function $V$ defined in
\eqref{modified} in the following sense: $L(\id_S) = V(S)$ for all
$S\subseteq\mathcal{N}$, where $\id_S$ denotes the indicator vector of $S$.

The optimization program \eqref{eq:non-strategic} extends naturally to such
a relaxation. We define:
\begin{equation}\tag{$P_c$}\label{eq:primal}
    L^*_c \defeq \max_{\lambda\in[0, 1]^{n}}
    \left\{L(\lambda) \Big| \sum_{i=1}^{n} \lambda_i c_i
    \leq B\right\}
\end{equation}

\begin{proposition}\label{prop:relaxation}
     $   L^*(c) \leq 2 OPT
        + 2\max_{i\in\mathcal{N}}V(i)$.
\end{proposition}

The proof of this proposition follows the \emph{pipage rounding} framework of
\citeN{pipage}.

This framework uses the \emph{multi-linear} extension $F$ of the submodular
function $V$. Let $P_\mathcal{N}^\lambda(S)$ be the probability of choosing the
set $S$ if we select each element $i$ in $\mathcal{N}$ independently with
probability $\lambda_i$:
\begin{displaymath}
    P_\mathcal{N}^\lambda(S) \defeq \prod_{i\in S} \lambda_i
    \prod_{i\in\mathcal{N}\setminus S}( 1 - \lambda_i).
\end{displaymath}
Then, the \emph{multi-linear} extension $F$ is defined by:
\begin{displaymath}
    F(\lambda) 
    \defeq \mathbb{E}_{S\sim P_\mathcal{N}^\lambda}\big[V(S)\big]
    = \sum_{S\subseteq\mathcal{N}} P_\mathcal{N}^\lambda(S) V(S)
\end{displaymath}

For \EDP{}  the multi-linear extension can be written:
\begin{equation}\label{eq:multi-linear-logdet}
    F(\lambda) = \mathbb{E}_{S\sim
    P_\mathcal{N}^\lambda}\bigg[\log\det \big(I_d + \sum_{i\in S} x_i\T{x_i}\big) \Big].
\end{equation}
Note that the relaxation $L$ that we introduced in \eqref{eq:our-relaxation},
follows naturally from the \emph{multi-linear} relaxation by swapping the
expectation and the $\log\det$ in \eqref{eq:multi-linear-logdet}:
\begin{displaymath}
    L(\lambda) = \log\det\left(\mathbb{E}_{S\sim
    P_\mathcal{N}^\lambda}\bigg[I_d + \sum_{i\in S} x_i\T{x_i} \bigg]\right).
\end{displaymath}

The proof proceeds as follows:
\begin{itemize}
\item First, we prove that $F$ admits the following rounding property: let
$\lambda$ be a feasible element of $[0,1]^n$, it is possible to trade one
fractional component of $\lambda$ for another until one of them becomes
integral, obtaining a new element $\tilde{\lambda}$ which is both feasible and
for which $F(\tilde{\lambda})\geq F(\lambda)$. Here, by feasibility of a point
$\lambda$, we mean that it satisfies the budget constraint $\sum_{i=1}^n
\lambda_i c_i \leq B$.  This rounding property is referred to in the literature
as \emph{cross-convexity} (see, \emph{e.g.}, \cite{dughmi}), or
$\varepsilon$-convexity by \citeN{pipage}. This is stated and proven in
Lemma~\ref{lemma:rounding} and allows us to bound $F$ in terms of $OPT$.
\item Next, we prove the central result of bounding $L$  appropriately in terms
of the multi-linear relaxation $F$ (Lemma \ref{lemma:relaxation-ratio}).
\item Finally, we conclude the proof of Proposition~\ref{prop:relaxation} by
combining Lemma~\ref{lemma:rounding} and Lemma~\ref{lemma:relaxation-ratio}.
\end{itemize}

\begin{lemma}[Rounding]\label{lemma:rounding}
    For any feasible $\lambda\in[0,1]^{n}$, there exists a feasible
    $\bar{\lambda}\in[0,1]^{n}$ such that at most one of its components is
    fractional %, that is, lies in $(0,1)$ and:
     and $F_{\mathcal{N}}(\lambda)\leq F_{\mathcal{N}}(\bar{\lambda})$.
\end{lemma}
\begin{proof}
    We give a rounding procedure which, given a feasible $\lambda$ with at least
    two fractional components, returns some feasible $\lambda'$ with one less fractional
    component such that $F(\lambda) \leq F(\lambda')$.

    Applying this procedure recursively yields the lemma's result.
    Let us consider such a feasible $\lambda$. Let $i$ and $j$ be two
    fractional components of $\lambda$ and let us define the following
    function:
    \begin{displaymath}
        F_\lambda(\varepsilon) = F(\lambda_\varepsilon)
        \quad\textrm{where} \quad
        \lambda_\varepsilon = \lambda + \varepsilon\left(e_i-\frac{c_i}{c_j}e_j\right)
    \end{displaymath}
    It is easy to see that if $\lambda$ is feasible, then:
    \begin{equation}\label{eq:convex-interval}
        \forall\varepsilon\in\Big[\max\Big(-\lambda_i,(\lambda_j-1)\frac{c_j}{c_i}\Big), \min\Big(1-\lambda_i, \lambda_j
        \frac{c_j}{c_i}\Big)\Big],\;
            \lambda_\varepsilon\;\;\textrm{is feasible}
    \end{equation}
    Furthermore, the function $F_\lambda$ is convex; indeed:
    \begin{align*}
        F_\lambda(\varepsilon)
        & = \mathbb{E}_{S'\sim P_{\mathcal{N}\setminus\{i,j\}}^\lambda(S')}\Big[
        (\lambda_i+\varepsilon)\Big(\lambda_j-\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{i,j\})\\
        & + (\lambda_i+\varepsilon)\Big(1-\lambda_j+\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{i\})
         + (1-\lambda_i-\varepsilon)\Big(\lambda_j-\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{j\})\\
         & + (1-\lambda_i-\varepsilon)\Big(1-\lambda_j+\varepsilon\frac{c_i}{c_j}\Big)V(S')\Big]
    \end{align*}
    Thus, $F_\lambda$ is a degree 2 polynomial whose dominant coefficient is:
    \begin{displaymath}
        \frac{c_i}{c_j}\mathbb{E}_{S'\sim
        P_{\mathcal{N}\setminus\{i,j\}}^\lambda(S')}\Big[
            V(S'\cup\{i\})+V(S'\cup\{i\})\\
        -V(S'\cup\{i,j\})-V(S')\Big]
    \end{displaymath}
    which is positive by submodularity of $V$. Hence, the maximum of
    $F_\lambda$ over the interval given in \eqref{eq:convex-interval} is
    attained at one of its limit, at which either the $i$-th or $j$-th component of
    $\lambda_\varepsilon$ becomes integral.
\end{proof}

\begin{lemma}\label{lemma:relaxation-ratio}
 %   The following inequality holds:
For all $\lambda\in[0,1]^{n},$
    %\begin{displaymath}
     $           \frac{1}{2}
        \,L(\lambda)\leq
        F(\lambda)\leq L(\lambda)$.
    %\end{displaymath}
\end{lemma}
\begin{proof}
    The bound $F_{\mathcal{N}}(\lambda)\leq L_{\mathcal{N}(\lambda)}$ follows by the concavity of the $\log\det$ function.
    To show the lower bound, 
    we first prove that $\frac{1}{2}$ is a lower bound of the ratio $\partial_i
    F(\lambda)/\partial_i L(\lambda)$, where
    $\partial_i\, \cdot$ denotes the partial derivative with respect to the
    $i$-th variable.  

   Let us start by computing the derivatives of $F$ and
    $L$ with respect to the $i$-th component.
    Observe that
    \begin{displaymath}
        \partial_i P_\mathcal{N}^\lambda(S) = \left\{
            \begin{aligned}
                & P_{\mathcal{N}\setminus\{i\}}^\lambda(S\setminus\{i\})\;\textrm{if}\;
                i\in S, \\
                & - P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\;\textrm{if}\;
                i\in \mathcal{N}\setminus S. \\
            \end{aligned}\right.
    \end{displaymath}
    Hence,
    \begin{displaymath}
        \partial_i F(\lambda) =
        \sum_{\substack{S\subseteq\mathcal{N}\\ i\in S}}
        P_{\mathcal{N}\setminus\{i\}}^\lambda(S\setminus\{i\})V(S)
        - \sum_{\substack{S\subseteq\mathcal{N}\\ i\in \mathcal{N}\setminus S}}
        P_{\mathcal{N}\setminus\{i\}}^\lambda(S)V(S).
    \end{displaymath}
    Now, using that every $S$ such that $i\in S$ can be uniquely written as
    $S'\cup\{i\}$, we can write:
    \begin{displaymath}
        \partial_i F(\lambda) =
        \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}}
        P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\big(V(S\cup\{i\})
        - V(S)\big).
    \end{displaymath}
    The marginal contribution of $i$ to
    $S$ can be written as 
\begin{align*}
V(S\cup \{i\}) - V(S)& =   \frac{1}{2}\log\det(I_d 
    + \T{X_S}X_S + x_i\T{x_i})
    - \frac{1}{2}\log\det(I_d + \T{X_S}X_S)\\
    &  = \frac{1}{2}\log\det(I_d + x_i\T{x_i}(I_d +
\T{X_S}X_S)^{-1})
 = \frac{1}{2}\log(1 + \T{x_i}A(S)^{-1}x_i)
\end{align*}
where $A(S) \defeq I_d+ \T{X_S}X_S$, and the last equality follows from the
Sylvester's determinant identity~\cite{sylvester}.
%  $  V(S\cup\{i\}) - V(S) = \frac{1}{2}\log\left(1 +  \T{x_i} A(S)^{-1}x_i\right)$.
Using this,
    \begin{displaymath}
        \partial_i F(\lambda) = \frac{1}{2}
        \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}}
        P_{\mathcal{N}\setminus\{i\}}^\lambda(S)
        \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big)
    \end{displaymath}
     The computation of the derivative of $L$ uses standard matrix
    calculus: writing $\tilde{A}(\lambda) = I_d+\sum_{i\in
    \mathcal{N}}\lambda_ix_i\T{x_i}$,
    \begin{displaymath}
        \det \tilde{A}(\lambda + h\cdot e_i) = \det\big(\tilde{A}(\lambda)
        + hx_i\T{x_i}\big)
         =\det \tilde{A}(\lambda)\big(1+
        h\T{x_i}\tilde{A}(\lambda)^{-1}x_i\big).
    \end{displaymath}
    Hence,
    \begin{displaymath}
       \log\det\tilde{A}(\lambda + h\cdot e_i)= \log\det\tilde{A}(\lambda)
        + h\T{x_i}\tilde{A}(\lambda)^{-1}x_i + o(h),
    \end{displaymath}
    which implies
    \begin{displaymath}
        \partial_i L(\lambda)
        =\frac{1}{2} \T{x_i}\tilde{A}(\lambda)^{-1}x_i.
    \end{displaymath}

For two symmetric matrices $A$ and $B$, we write $A\succ B$ ($A\succeq B$) if
$A-B$ is positive definite (positive semi-definite).  This order allows us to
define the notion of a \emph{decreasing} as well as  \emph{convex} matrix
function, similarly to their real counterparts. With this definition, matrix
inversion is decreasing and convex over symmetric positive definite
matrices (see Example 3.48 p. 110 in \cite{boyd2004convex}).
In particular,
\begin{gather*}
        \forall S\subseteq\mathcal{N},\quad A(S)^{-1} \succeq A(S\cup\{i\})^{-1}
\end{gather*}
as $A(S)\preceq A(S\cup\{i\})$. Observe that
 %   \begin{gather*}
 %       \forall 
$P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\geq P_{\mathcal{N}\setminus\{i\}}^\lambda(S\cup\{i\})$ for all $S\subseteq\mathcal{N}\setminus\{i\}$, and
       % ,\\
        $P_{\mathcal{N}\setminus\{i\}}^\lambda(S) \geq P_\mathcal{N}^\lambda(S),$ for all $S\subseteq\mathcal{N}$.
 %\end{gather*}
    Hence,
    \begin{align*}
        \partial_i F(\lambda) 
      %  & = \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} P_\mathcal{N}^\lambda(S) \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big)\\
        & \geq \frac{1}{4}
        \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}}
        P_{\mathcal{N}\setminus\{i\}}^\lambda(S)
        \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big)\\
        &\hspace{-3.5em}+\frac{1}{4}
        \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}}
        P_{\mathcal{N}\setminus\{i\}}^\lambda(S\cup\{i\})
        \log\Big(1 + \T{x_i}A(S\cup\{i\})^{-1}x_i\Big)\\
        &\geq \frac{1}{4}
        \sum_{S\subseteq\mathcal{N}}
        P_\mathcal{N}^\lambda(S)
        \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big).
    \end{align*}
    Using that $A(S)\succeq I_d$ we get that $\T{x_i}A(S)^{-1}x_i \leq
    \norm{x_i}_2^2 \leq 1$. Moreover, $\log(1+x)\geq x$ for all $x\leq 1$.
    Hence,
    \begin{displaymath}
        \partial_i F(\lambda) \geq
        \frac{1}{4}
        \T{x_i}\bigg(\sum_{S\subseteq\mathcal{N}}P_\mathcal{N}^\lambda(S)A(S)^{-1}\bigg)x_i.
    \end{displaymath}
    Finally, using that the inverse is a matrix convex function over symmetric
    positive definite matrices:
    \begin{displaymath}
        \partial_i F(\lambda) \geq
        \frac{1}{4}
        \T{x_i}\bigg(\sum_{S\subseteq\mathcal{N}}P_\mathcal{N}^\lambda(S)A(S)\bigg)^{-1}x_i
        = \frac{1}{4}\T{x_i}\tilde{A}(\lambda)^{-1}x_i
         = \frac{1}{2}
     \partial_i L(\lambda).
    \end{displaymath}

Having bound the ratio between the partial derivatives, we now bound the ratio $F(\lambda)/L(\lambda)$ from below. Consider the following cases.
    First, if the minimum of the ratio
    $F(\lambda)/L(\lambda)$ is attained at a point interior to the hypercube, then it is
    a critical point, \emph{i.e.}, $\partial_i \big(F(\lambda)/L(\lambda)\big)=0$ for all $i\in \mathcal{N}$; hence, at such a critical point:
    \begin{equation}\label{eq:lhopital}
        \frac{F(\lambda)}{L(\lambda)}
        = \frac{\partial_i F(\lambda)}{\partial_i
        L(\lambda)} \geq \frac{1}{2}.
    \end{equation}
    Second, if the minimum is attained as 
    $\lambda$ converges to zero in, \emph{e.g.}, the $l_2$ norm, by the Taylor approximation, one can write:
    \begin{displaymath}
        \frac{F(\lambda)}{L(\lambda)}
        \sim_{\lambda\rightarrow 0}
        \frac{\sum_{i\in \mathcal{N}}\lambda_i\partial_i F(0)}
        {\sum_{i\in\mathcal{N}}\lambda_i\partial_i L(0)}
        \geq \frac{1}{2},
    \end{displaymath}
    \emph{i.e.}, the ratio $\frac{F(\lambda)}{L(\lambda)}$ is necessarily bounded from below by 1/2 for small enough $\lambda$. 
       Finally, if the minimum is attained on a face of the hypercube $[0,1]^n$ (a face is
    defined as a subset of the hypercube where one of the variable is fixed to
    0 or 1), without loss of generality, we can assume that the minimum is
    attained on the face where the $n$-th variable has been fixed
    to 0 or 1. Then, either the minimum is attained at a point interior to the
    face or on a boundary of the face. In the first sub-case, relation
    \eqref{eq:lhopital} still characterizes the minimum for $i< n$.
    In the second sub-case, by repeating the argument again by induction, we see
    that all is left to do is to show that the bound holds for the vertices of
    the cube (the faces of dimension 1).  The vertices are exactly the binary
    points, for which we know that both relaxations are equal to the value
    function $V$. Hence, the ratio is equal to 1 on the vertices.
\end{proof}

To conclude the proof of Proposition~\ref{prop:relaxation}, let us consider
a feasible point $\lambda^*\in[0,1]^{n}$ such that $L(\lambda^*) = L^*_c$. By
applying Lemma~\ref{lemma:relaxation-ratio} and Lemma~\ref{lemma:rounding} we
get a feasible point $\bar{\lambda}$ with at most one fractional component such
that
\begin{equation}\label{eq:e1}
    L(\lambda^*) \leq 2 F(\bar{\lambda}).
\end{equation}
    Let $\lambda_i$ denote the fractional component of $\bar{\lambda}$ and $S$
    denote the set whose indicator vector is $\bar{\lambda} - \lambda_i e_i$.
    By definition of the multi-linear extension $F$:
    \begin{displaymath}
        F(\bar{\lambda}) = (1-\lambda_i)V(S) +\lambda_i V(S\cup\{i\}).
    \end{displaymath}
    By submodularity of $V$, $V(S\cup\{i\})\leq V(S) + V(\{i\})$. Hence,
    \begin{displaymath}
        F(\bar{\lambda}) \leq V(S) + V(i).
    \end{displaymath}
    Note that since $\bar{\lambda}$ is feasible, $S$ is also feasible and
    $V(S)\leq OPT$. Hence,
    \begin{equation}\label{eq:e2}
        F(\bar{\lambda}) \leq  OPT + \max_{i\in\mathcal{N}} V(i).
    \end{equation}
    Together, \eqref{eq:e1} and \eqref{eq:e2} imply the lemma. \hspace*{\stretch{1}}\qed

\subsection{A monotonous Newton's estimator}\label{sec:monotonicity}

\textbf{TODO} Explain that we only get approximate monotonicity, but even
that is not immediate since the variation induced by a change of cost on
coordinate $i$ depends on the allocation at this coordinate which can be
arbitrarily small.

For the ease of presentation, we normalize the costs by dividing them by the
budget $B$ so that the budget constraint in \eqref{eq:primal} now reads
$\T{c}\lambda\leq 1$.

\begin{proposition}\label{prop:monotonicity}
    Let $\delta\in(0,1]$. For any $\varepsilon\in(0,1]$, there exists
    an algorithm which computes an approximate solution $\tilde{L}^*_c$ to
    \eqref{eq:primal} such that:
    \begin{enumerate}
        \item $|\tilde{L}^*_c - L^*_c| \leq \varepsilon$
        \item for all $c' = (c_i', c_{-i})$ with $c_i'\leq c_i-\delta$, $\tilde{L}^*_c \leq \tilde{L}^*_{c'}$
        \item the routine's running time is $O\big(poly(n, d, \log\log\frac{1}{b\varepsilon\delta})\big)$
    \end{enumerate}
\end{proposition}

We consider a perturbation of \eqref{eq:primal} by introducing:
\begin{equation}\tag{$P_{c, \alpha}$}\label{eq:perturbed-primal}
    L^*_c(\alpha) \defeq \max_{\lambda\in[\alpha, 1]^{n}}
    \left\{L(\lambda) \Big| \sum_{i=1}^{n} \lambda_i c_i
    \leq B\right\}
\end{equation}
Note that we have $L^*_c = L^*_c(0)$.  We will also assume that
$\alpha<\frac{1}{n}$ so that \eqref{eq:perturbed-primal} has at least one
feasible point: $(\frac{1}{n},\ldots,\frac{1}{n})$.

Having introduced this perturbed problem, we show that its optimal value is
close to the optimal value of \eqref{eq:primal} (Lemma~\ref{lemma:proximity})
while being well-behaved with respect to changes of the cost
(Lemma~\ref{lemma:monotonicity}). These lemmas together imply
Proposition~\ref{prop:monotonicity}.


\begin{lemma}\label{lemma:derivative-bounds}
    Let $\partial_i L(\lambda)$ denote the $i$-th derivative of $L$, for $i\in\{1,\ldots, n\}$, then:
    \begin{displaymath}
        \forall\lambda\in[0, 1]^n,\;\frac{b}{2^n} \leq \partial_i L(\lambda) \leq 1
    \end{displaymath}
\end{lemma}

\begin{proof}
    Let us define:
    \begin{displaymath}
        S(\lambda)\defeq I_d + \sum_{i=1}^n \lambda_i x_i\T{x_i}
        \quad\mathrm{and}\quad
        S_k \defeq I_d + \sum_{i=1}^k x_i\T{x_i}
    \end{displaymath}

    We have $\partial_i L(\lambda) = \T{x_i}S(\lambda)^{-1}x_i$. Since
    $S(\lambda)\geq I_d$, $\partial_i L(\lambda)\leq \T{x_i}x_i \leq 1$, which
    is the right-hand side  of the lemma.

    For the left-hand side, note that $S(\lambda) \leq S_n$. Hence
    $\partial_iL(\lambda)\geq \T{x_i}S_n^{-1}x_i$.

    Using the Sherman-Morrison formula, for all $k\geq 1$:
    \begin{displaymath}
        \T{x_i}S_k^{-1} x_i = \T{x_i}S_{k-1}^{-1}x_i 
        - \frac{(\T{x_i}S_{k-1}^{-1}x_k)^2}{1+\T{x_k}S_{k-1}^{-1}x_k}
    \end{displaymath}

    By the Cauchy-Schwarz inequality:
    \begin{displaymath}
        (\T{x_i}S_{k-1}^{-1}x_k)^2 \leq \T{x_i}S_{k-1}^{-1}x_i\;\T{x_k}S_{k-1}^{-1}x_k
    \end{displaymath}

    Hence:
    \begin{displaymath}
        \T{x_i}S_k^{-1} x_i \geq \T{x_i}S_{k-1}^{-1}x_i 
        - \T{x_i}S_{k-1}^{-1}x_i\frac{\T{x_k}S_{k-1}^{-1}x_k}{1+\T{x_k}S_{k-1}^{-1}x_k}
    \end{displaymath}

    But $\T{x_k}S_{k-1}^{-1}x_k\leq 1$ and $\frac{a}{1+a}\leq \frac{1}{2}$ if
    $0\leq a\leq 1$, so:
    \begin{displaymath}
        \T{x_i}S_{k}^{-1}x_i \geq \T{x_i}S_{k-1}^{-1}x_i
        - \frac{1}{2}\T{x_i}S_{k-1}^{-1}x_i\geq \frac{\T{x_i}S_{k-1}^{-1}x_i}{2}
    \end{displaymath}

    By induction:
    \begin{displaymath}
        \T{x_i}S_n^{-1} x_i \geq \frac{\T{x_i}x_i}{2^n}
    \end{displaymath}
    
    Using that $\T{x_i}{x_i}\geq b$ concludes the proof of the left-hand side
    of the lemma's inequality.
\end{proof}

Let us introduce the lagrangian of problem, \eqref{eq:perturbed-primal}:

\begin{displaymath}
    \mathcal{L}_{c, \alpha}(\lambda, \mu, \nu, \xi) \defeq L(\lambda) 
    + \T{\mu}(\lambda-\alpha\mathbf{1}) + \T{\nu}(\mathbf{1}-\lambda) + \xi(1-\T{c}\lambda)
\end{displaymath}
so that:
\begin{displaymath}
    L^*_c(\alpha) = \min_{\mu, \nu, \xi\geq 0}\max_\lambda \mathcal{L}_{c, \alpha}(\lambda, \mu, \nu, \xi)
\end{displaymath}
Similarly, we define $\mathcal{L}_{c}\defeq\mathcal{L}_{c, 0}$ the lagrangian of \eqref{eq:primal}.

Let $\lambda^*$ be primal optimal for \eqref{eq:perturbed-primal}, and $(\mu^*,
\nu^*, \xi^*)$ be dual optimal for the same problem. In addition to primal and
dual feasibility, the KKT conditions give $\forall i\in\{1, \ldots, n\}$:
\begin{gather*}
    \partial_i L(\lambda^*) + \mu_i^* - \nu_i^* - \xi^* c_i = 0\\
    \mu_i^*(\lambda_i^* - \alpha) = 0\\
    \nu_i^*(1 - \lambda_i^*) = 0
\end{gather*}

\begin{lemma}\label{lemma:proximity}
We have:
\begin{displaymath}
    L^*_c - \alpha n^2\leq L^*_c(\alpha) \leq L^*_c
\end{displaymath}
In particular, $|L^*_c - L^*_c(\alpha)| \leq \alpha n^2$.
\end{lemma}

\begin{proof}
    $\alpha\mapsto L^*_c(\alpha)$ is a decreasing function as it is the
    maximum value of the $L$ function over a set-decreasing domain, which gives
    the rightmost inequality.

    Let $\mu^*, \nu^*, \xi^*$ be dual optimal for $(P_{c, \alpha})$, that is:
    \begin{displaymath}
        L^*_{c}(\alpha) = \max_\lambda \mathcal{L}_{c, \alpha}(\lambda, \mu^*, \nu^*, \xi^*)
    \end{displaymath}

    Note that $\mathcal{L}_{c, \alpha}(\lambda, \mu^*, \nu^*, \xi^*)
    = \mathcal{L}_{c}(\lambda, \mu^*, \nu^*, \xi^*)
    - \alpha\T{\mathbf{1}}\mu^*$, and that for any $\lambda$ feasible for
    problem \eqref{eq:primal}, $\mathcal{L}_{c}(\lambda, \mu^*, \nu^*, \xi^*)
    \geq L(\lambda)$. Hence,
    \begin{displaymath}
        L^*_{c}(\alpha) \geq L(\lambda) - \alpha\T{\mathbf{1}}\mu^*
    \end{displaymath}
    for any $\lambda$ feasible for \eqref{eq:primal}. In particular, for $\lambda$ primal optimal for $\eqref{eq:primal}$:
    \begin{equation}\label{eq:local-1}
        L^*_{c}(\alpha) \geq L^*_c - \alpha\T{\mathbf{1}}\mu^*
    \end{equation}

    Let us denote by the $M$ the support of $\mu^*$, that is $M\defeq
    \{i|\mu_i^* > 0\}$, and by $\lambda^*$ a primal optimal point for
    $\eqref{eq:perturbed-primal}$.  From the KKT conditions we see that:
    \begin{displaymath}
        M \subseteq \{i|\lambda_i^* = \alpha\}
    \end{displaymath}


    Let us first assume that $|M| = 0$, then $\T{\mathbf{1}}\mu^*=0$ and the lemma follows.

    We will now assume that $|M|\geq 1$. In this case $\T{c}\lambda^*
    = 1$, otherwise we could increase the coordinates of $\lambda^*$ in $M$,
    which would increase the value of the objective function and contradict the
    optimality of $\lambda^*$. Note also, that $|M|\leq n-1$, otherwise, since
    $\alpha< \frac{1}{n}$, we would have $\T{c}\lambda^*\ < 1$, which again
    contradicts the optimality of $\lambda^*$. Let us write:
    \begin{displaymath}
        1 = \T{c}\lambda^* = \alpha\sum_{i\in M}c_i + \sum_{i\in \bar{M}}\lambda_i^*c_i
        \leq \alpha |M| + (n-|M|)\max_{i\in \bar{M}} c_i
    \end{displaymath}
    That is:
    \begin{equation}\label{local-2}
        \max_{i\in\bar{M}} c_i \geq \frac{1 - |M|\alpha}{n-|M|}> \frac{1}{n}
    \end{equation}
    where the last inequality uses again that $\alpha<\frac{1}{n}$. From the
    KKT conditions, we see that for $i\in M$, $\nu_i^* = 0$ and:
    \begin{equation}\label{local-3}
        \mu_i^* = \xi^*c_i - \partial_i L(\lambda^*)\leq \xi^*c_i\leq \xi^*
    \end{equation}
    since $\partial_i L(\lambda^*)\geq 0$ and $c_i\leq 1$.

    Furthermore, using the KKT conditions again, we have that:
    \begin{equation}\label{local-4}
        \xi^* \leq \inf_{i\in \bar{M}}\frac{\partial_i L(\lambda^*)}{c_i}\leq \inf_{i\in \bar{M}} \frac{1}{c_i}
        = \frac{1}{\max_{i\in\bar{M}} c_i}
    \end{equation}
    where the last inequality uses Lemma~\ref{lemma:derivative-bounds}.

    Combining \eqref{local-2}, \eqref{local-3} and \eqref{local-4}, we get that:
    \begin{displaymath}
        \sum_{i\in M}\mu_i^* \leq |M|\xi^* \leq n\xi^*\leq \frac{n}{\max_{i\in\bar{M}} c_i} \leq n^2
    \end{displaymath}
    
    This implies that:
    \begin{displaymath}
        \T{\mathbf{1}}\mu^* = \sum_{i=1}^n \mu^*_i = \sum_{i\in M}\mu_i^*\leq n^2
    \end{displaymath}
    which in addition to \eqref{eq:local-1} proves the lemma.
\end{proof} 

\begin{lemma}\label{lemma:monotonicity}
    If $c'$ = $(c_i', c_{-i})$, with $c_i'\leq c_i - \delta$, we have:
    \begin{displaymath}
        L^*_{c'}(\alpha) \geq L^*_c(\alpha) + \frac{\alpha\delta b}{2^n}
    \end{displaymath}
\end{lemma}

\begin{proof}
    Let $\mu^*, \nu^*, \xi^*$ be dual optimal for $(P_{c', \alpha})$. Noting that:
    \begin{displaymath} 
    \mathcal{L}_{c', \alpha}(\lambda, \mu^*, \nu^*, \xi^*) \geq
    \mathcal{L}_{c, \alpha}(\lambda, \mu^*, \nu^*, \xi^*) + \lambda_i\xi^*\delta,
    \end{displaymath}
    we get similarly to Lemma~\ref{lemma:proximity}:
    \begin{displaymath}
        L^*_{c'}(\alpha) \geq L(\lambda) + \lambda_i\xi^*\delta
    \end{displaymath}
    for any $\lambda$ feasible for \eqref{eq:perturbed-primal}. In particular, for $\lambda^*$ primal optimal for \eqref{eq:perturbed-primal}:
    \begin{displaymath}
        L^*_{c'}(\alpha) \geq L^*_{c}(\alpha) + \alpha\xi^*\delta
    \end{displaymath}
    since $\lambda_i^*\geq \alpha$.

    Using the KKT conditions for $(P_{c', \alpha})$, we can write:
    \begin{displaymath}
        \xi^* = \inf_{i:\lambda^{'*}_i>\alpha} \frac{\T{x_i}S(\lambda^{'*})^{-1}x_i}{c_i'}
    \end{displaymath}
    with $\lambda^{'*}$ optimal for $(P_{c', \alpha})$. Since $c_i'\leq 1$, using Lemma~\ref{lemma:derivative-bounds}, we get that $\xi^*\geq \frac{b}{2^n}$, which concludes the proof.
\end{proof}


\subsubsection*{End of the proof of Proposition~\ref{prop:monotonicity}}

Let $\varepsilon$ in $(0, 1]$. The routine works as follows: set $\alpha\defeq
\varepsilon(\delta + n^2)^{-1}$ and return an approximation $\tilde{L}^*_c$ of
$L^*_c(\alpha)$ with an accuracy $\frac{1}{2^{n+1}}\alpha\delta b$ computed by
a standard convex optimization algorithm. Note that this choice of $\alpha$
implies $\alpha<\frac{1}{n}$ as required.

\begin{enumerate}
    \item using Lemma~\ref{lemma:proximity}:
\begin{displaymath}
        |\tilde{L}^*_c - L^*_c| \leq |\tilde{L}^*_c - L^*_c(\alpha)| + |L^*_c(\alpha) - L^*_c|
        \leq \alpha\delta + \alpha n^2 = \varepsilon
\end{displaymath}

\item let $c' = (c_i', c_{-i})$ with $c_i'\leq c_i-\delta$, then:
\begin{displaymath}
    \tilde{L}^*_{c'} \geq L^*_{c'} - \frac{\alpha\delta b}{2^{n+1}} 
                     \geq L^*_c + \frac{\alpha\delta b}{2^{n+1}}
    \geq \tilde{L}^*_c
\end{displaymath}
where the first and inequality come from the accuracy of the approximation, and
the inner inequality follows from Lemma~\ref{lemma:monotonicity}.

\item the accuracy of the approximation $\tilde{L}^*_c$ is:
\begin{displaymath}
    A\defeq\frac{\varepsilon\delta b}{2^{n+1}(\delta + n^2)}
\end{displaymath}

The function $L$ is well-known to be concave and even self-concordant (see
\emph{e.g.}, \cite{boyd2004convex}).  In this case, the analysis of Newton's
method for self-concordant functions in \cite{boyd2004convex}, shows that
finding the maximum of $L$ to any precision $A$ can be done in
$O(\log\log A^{-1})$ iterations. Note that:
\begin{displaymath}
    \log\log A^{-1} = O\bigg(\log\log\frac{1}{\varepsilon\delta b} + \log n\bigg)
\end{displaymath}
Furthermore, each iteration of Newton's method can be done in time $O\big(poly(n,
d)\big)$.\qed
\end{enumerate}