
\section{Proofs of Statements in Section~\ref{sec:concave}}
\subsection{Proof of Lemma~\ref{lemma:relaxation-ratio}}\label{proofofrelaxation-ratio}
%\begin{proof}
    The bound $F_{\mathcal{N}}(\lambda)\leq L_{\mathcal{N}}(\lambda)$ follows by the concavity of the $\log\det$ function and Jensen's inequality.
    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 we use
    $\partial_i\, \cdot$ as a shorthand for  $\frac{\partial}{\partial \lambda_i}$, the partial derivative  with respect to the
    $i$-th variable. 

   Let us start by computing the partial 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 since $1\leq \lambda_i\leq 1$, 
$P_{\mathcal{N}\setminus\{i\}}^\lambda(S) \geq P_\mathcal{N}^\lambda(S)$ and
$P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\geq P_{\mathcal{N}}^\lambda(S\cup\{i\})$
for all $S\subseteq\mathcal{N}\setminus\{i\}$. Hence,
\begin{align*}
    \partial_i F(\lambda) 
    & \geq \frac{1}{4}
    \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)\\
    &\hspace{-3.5em}+\frac{1}{4}
    \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}}
    P_{\mathcal{N}}^\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 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 $\lambda$.

Second, if the minimum of the ratio $F(\lambda)/L(\lambda)$ is attained at
a vertex of the hypercube $[0,1]^n$ different from 0. $F$ and $L$ being
relaxations of the value function $V$, they are equal to $V$ on the vertices
which are exactly the binary points. Hence, the minimum is equal to 1 in this
case; in particular, it is greater than $1/2$.

Finally, if the minimum is attained at a point $\lambda^*$ with at least one
coordinate belonging to $(0,1)$, let $i$ be one such coordinate and consider
the function $G_i$:
\begin{displaymath}
    G_i: x \mapsto \frac{F}{L}(\lambda_1^*,\ldots,\lambda_{i-1}^*, x,
    \lambda_{i+1}^*, \ldots, \lambda_n^*).
\end{displaymath}
Then this function attains a minimum at $\lambda^*_i\in(0,1)$ and its
derivative is zero at this point. Hence:
\begin{displaymath}
    0 = G_i'(\lambda^*_i) = \partial_i\left(\frac{F}{L}\right)(\lambda^*).
\end{displaymath}
But $\partial_i(F/L)(\lambda^*)=0$ implies that
\begin{displaymath}
    \frac{F(\lambda^*)}{L(\lambda^*)} = \frac{\partial_i
    F(\lambda^*)}{\partial_i L(\lambda^*)}\geq \frac{1}{2}
\end{displaymath}
using the lower bound on the ratio of the partial derivatives. This concludes
the proof of the lemma. \qed
%\end{proof}

%We now prove that $F$ admits the following exchange 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}.

%\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}
\subsection{Proof of Lemma~\ref{lemma:rounding}}\label{proofoflemmarounding}
%\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. \qed
%\end{proof}
\subsection{Proof of Proposition~\ref{prop:relaxation}}\label{proofofproprelaxation}
The lower bound on $L^*_c$ follows immediately from the fact that $L$ extends $V$ to $[0,1]^n$. For the upper bound, let us consider a feasible point $\lambda^*\in \dom_c$ 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 proposition.\qed

\section{Proof of Proposition~\ref{prop:monotonicity}}\label{proofofpropmonotonicity}

%The $\log\det$ function is concave and self-concordant (see
%\cite{boyd2004convex}), in this case, the analysis of the barrier method in
%in \cite{boyd2004convex} (Section 11.5.5) can be summarized in the following
%lemma:

%\begin{lemma}\label{lemma:barrier}
%For any $\varepsilon>0$, the barrier method computes an $\varepsilon$-accurate
%approximation of $L^*_c$ in time $O(poly(n,d,\log\log\varepsilon^{-1})$.
%\end{lemma}
\note{THIBAUT: ARE THE COSTS BELOW DIVIDED BY B? CAN YOU EITHER RENAME THEM OR CARRY THE B AROUND AS APPROPRIATE?}

We proceed by showing that the optimal value of \eqref{eq:perturbed-primal} 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}
    Recall that we had defined:
    \begin{displaymath}
        \tilde{A}(\lambda)\defeq I_d + \sum_{i=1}^n \lambda_i x_i\T{x_i}
        \quad\mathrm{and}\quad
        A(S) \defeq I_d + \sum_{i\in S} x_i\T{x_i}
    \end{displaymath}
    Let us also define $A_k\defeq A(\{x_1,\ldots,x_k\})$.
    We have $\partial_i L(\lambda) = \T{x_i}\tilde{A}(\lambda)^{-1}x_i$. Since
    $\tilde{A}(\lambda)\succeq I_d$, $\partial_i L(\lambda)\leq \T{x_i}x_i \leq 1$, which
    is the right-hand side  of the lemma.
    For the left-hand side, note that $\tilde{A}(\lambda) \preceq A_n$. Hence
    $\partial_iL(\lambda)\geq \T{x_i}A_n^{-1}x_i$.
    Using the Sherman-Morrison formula, for all $k\geq 1$:
    \begin{displaymath}
        \T{x_i}A_k^{-1} x_i = \T{x_i}A_{k-1}^{-1}x_i 
        - \frac{(\T{x_i}A_{k-1}^{-1}x_k)^2}{1+\T{x_k}A_{k-1}^{-1}x_k}
    \end{displaymath}
    By the Cauchy-Schwarz inequality:
    \begin{displaymath}
        (\T{x_i}A_{k-1}^{-1}x_k)^2 \leq \T{x_i}A_{k-1}^{-1}x_i\;\T{x_k}A_{k-1}^{-1}x_k
    \end{displaymath}
    Hence:
    \begin{displaymath}
        \T{x_i}A_k^{-1} x_i \geq \T{x_i}A_{k-1}^{-1}x_i 
        - \T{x_i}A_{k-1}^{-1}x_i\frac{\T{x_k}A_{k-1}^{-1}x_k}{1+\T{x_k}A_{k-1}^{-1}x_k}
    \end{displaymath}
    But $\T{x_k}A_{k-1}^{-1}x_k\leq 1$ and $\frac{a}{1+a}\leq \frac{1}{2}$ if
    $0\leq a\leq 1$, so:
    \begin{displaymath}
        \T{x_i}A_{k}^{-1}x_i \geq \T{x_i}A_{k-1}^{-1}x_i
        - \frac{1}{2}\T{x_i}A_{k-1}^{-1}x_i\geq \frac{\T{x_i}A_{k-1}^{-1}x_i}{2}
    \end{displaymath}
    By induction:
    \begin{displaymath}
        \T{x_i}A_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}

%\subsection*{End of the proof of Proposition~\ref{prop:monotonicity}}
We are now ready to conclude the  proof of Proposition~\ref{prop:monotonicity}.
Let $\hat{L}^*_{c,\alpha}$ be the approximation computed by
Algorithm~\ref{alg:monotone}.
\begin{enumerate}
    \item using Lemma~\ref{lemma:proximity}:
\begin{displaymath}
        |\hat{L}^*_{c,\alpha} - L^*_c| \leq |\hat{L}^*_{c,\alpha} - L^*_{c,\alpha}| + |L^*_{c,\alpha} - L^*_c|
        \leq \alpha\delta + \alpha n^2 = \varepsilon
\end{displaymath}
which proves the $\varepsilon$-accuracy.

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

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

Note that:
\begin{displaymath}
    \log\log A^{-1} = O\bigg(\log\log\frac{1}{\varepsilon\delta b} + \log n\bigg)
\end{displaymath}
Using Lemma~\ref{lemma:barrier} concludes the proof of the running time.\qed
\end{enumerate}

\section{Proof of Theorem~\ref{thm:main}}\label{sec:proofofmainthm}

We now present the proof of Theorem~\ref{thm:main}. $\delta$-truthfulness and
individual rationality follow from $\delta$-monotonicity and threshold
payments. $\delta$-monotonicity and budget feasibility follow the same steps as the
analysis of \citeN{chen}; for the sake of completeness, we restate their proof
here.

\begin{lemma}\label{lemma:monotone}
Our mechanism for \EDP{} is $\delta$-monotone and budget feasible.
\end{lemma}

\begin{proof}
    Consider an agent $i$ with cost $c_i$ that is
    selected by the mechanism, and suppose that she reports
    a cost $c_i'\leq c_i-\delta$ while all  other costs stay the same.
    Suppose that when $i$ reports $c_i$, $OPT'_{-i^*} \geq C V(i^*)$; then, as $s_i(c_i,c_{-i})=1$, $i\in S_G$.
     By reporting cost $c_i'$, $i$ may be selected at an earlier iteration of the greedy algorithm.
    %using the submodularity of $V$, we see that $i$ will satisfy the greedy
    %selection rule:
    %\begin{displaymath}
    %    i = \argmax_{j\in\mathcal{N}\setminus S} \frac{V(S\cup\{j\})
    %    - V(S)}{c_j}
    %\end{displaymath}
    %in an earlier iteration of the greedy heuristic.
 Denote by $S_i$
    (resp. $S_i'$) the set to which $i$ is added when reporting cost $c_i$
    (resp. $c_i'$). We have $S_i'\subseteq S_i$; in addition, $S_i'\subseteq S_G'$, the set selected by the greedy algorithm under $(c_i',c_{-i})$; if not, then greedy selection would terminate prior to selecting $i$ also when she reports $c_i$, a contradiction. Moreover, we have
    \begin{align*}
        c_i' & \leq c_i \leq
        \frac{B}{2}\frac{V(S_i\cup\{i\})-V(S_i)}{V(S_i\cup\{i\})}
         \leq \frac{B}{2}\frac{V(S_i'\cup\{i\})-V(S_i')}{V(S_i'\cup\{i\})}
    \end{align*}
    by the monotonicity and submodularity  of $V$. Hence  $i\in S_G'$. By
    $\delta$-decreasingness of
    $OPT'_{-i^*}$, under $c'_i\leq c_i-\delta$ the greedy set is still allocated and $s_i(c_i',c_{-i}) =1$.
    Suppose now that when $i$ reports $c_i$, $OPT'_{-i^*} < C V(i^*)$. Then $s_i(c_i,c_{-i})=1$ iff $i = i^*$. 
    Reporting $c_{i^*}'\leq c_{i^*}$ does not change $V(i^*)$ nor
    $OPT'_{-i^*} \leq C V(i^*)$; thus $s_{i^*}(c_{i^*}',c_{-i^*})=1$, so the mechanism is monotone.

To show budget feasibility, suppose that $OPT'_{-i^*} < C V(i^*)$. Then the mechanism selects $i^*$. Since the bid of $i^*$ does not affect the above condition, the threshold payment of $i^*$ is $B$ and the mechanism is budget feasible.
Suppose  that $OPT'_{-i^*} \geq C V(i^*)$. 
Denote by $S_G$ the set selected by the greedy algorithm, and for $i\in S_G$,  denote by
$S_i$ the subset of the solution set that was selected by the greedy algorithm just prior to the addition of $i$---both sets determined for the present cost vector $c$. 
%Chen \emph{et al.}~\cite{chen} show that, 
Then for any submodular function $V$, and for all $i\in S_G$:
%the reported cost of an agent selected by the greedy heuristic, and holds for
%any submodular function $V$:
\begin{equation}\label{eq:budget}
   \text{if}~c_i'\geq \frac{V(S_i\cup\{i\}) - V(S)}{V(S_G)} B~\text{then}~s_i(c_i',c_{-i})=0
\end{equation}
In other words, if $i$ increases her cost to a value higher than $\frac{V(S_i\cup\{i\}) - V(S)}{V(S_G)}$, she will cease to be in the selected set $S_G$. As a result,
\eqref{eq:budget}
implies that the threshold payment of user $i$ is bounded by the above quantity.
%\begin{displaymath}
%\frac{V(S_i\cup\{i\}) - V(S_i)}{V(S_G)} =  B
%\end{displaymath}
Hence, the total payment is bounded by the telescopic sum:
\begin{displaymath}
    \sum_{i\in S_G} \frac{V(S_i\cup\{i\}) - V(S_i)}{V(S_G)} B = \frac{V(S_G)-V(\emptyset)}{V(S_G)} B=B\qed
\end{displaymath}
\end{proof}

The complexity of the mechanism is given by the following lemma.

\begin{lemma}[Complexity]\label{lemma:complexity}
    For any $\varepsilon > 0$ and any $\delta>0$, the complexity of the mechanism  is
    $O\big(poly(n, d, \log\log\frac{1}{b\varepsilon\delta})\big)$
\end{lemma}

\begin{proof}
    The value function $V$ in \eqref{modified} can be computed in time
    $O(\text{poly}(n, d))$ and the mechanism only involves a linear
    number of queries to the function $V$. 

    By Proposition~\ref{prop:monotonicity}, line 3 of Algorithm~\ref{mechanism}
    can be computed in time
    $O(\text{poly}(n, d, \log\log \varepsilon^{-1}))$. Hence the allocation
    function's complexity is as stated. 
    %Payments can be easily computed in time  $O(\text{poly}(n, d))$ as in prior work.
\junk{
    Using Singer's characterization of the threshold payments
    \cite{singer-mechanisms}, one can verify that they can be computed in time
    $O(\text{poly}(n, d))$.
    }
\end{proof}

Finally, we prove the approximation ratio of the mechanism.

We use the following lemma from \cite{chen} which bounds $OPT$ in terms of
the value of $S_G$, as computed in Algorithm \ref{mechanism}, and $i^*$, the
element of maximum value.

\begin{lemma}[\cite{chen}]\label{lemma:greedy-bound}
Let $S_G$ be the set computed in Algorithm \ref{mechanism} and let 
$i^*=\argmax_{i\in\mathcal{N}} V(\{i\})$. We have:
\begin{displaymath}
OPT \leq \frac{e}{e-1}\big( 3 V(S_G) + 2 V(i^*)\big).
\end{displaymath}
\end{lemma}

Using Proposition~\ref{prop:relaxation} and~\ref{lemma:greedy-bound} we can complete the proof of
Theorem~\ref{thm:main} by showing that, for any $\varepsilon > 0$, if
$OPT_{-i}'$, the optimal value of $L$ when $i^*$ is excluded from
$\mathcal{N}$, has been computed to a precision $\varepsilon$, then the set
$S^*$ allocated by the mechanism is such that:
\begin{equation} \label{approxbound}
OPT
\leq \frac{10e\!-\!3 + \sqrt{64e^2\!-\!24e\!+\!9}}{2(e\!-\!1)} V(S^*)\!
+ \! \varepsilon .
\end{equation}
To see this, let $L^*_{-i^*}$ be the true maximum value of $L$ subject to
$\lambda_{i^*}=0$, $\sum_{i\in \mathcal{N}\setminus{i^*}}c_i\leq B$. From line
3 of Algorithm~\ref{mechanism}, we have
$L^*_{-i^*}-\varepsilon\leq OPT_{-i^*}' \leq L^*_{-i^*}+\varepsilon$.

If the condition on line 4 of the algorithm holds, then
\begin{displaymath}
    V(i^*) \geq \frac{1}{C}L^*_{-i^*}-\frac{\varepsilon}{C} \geq
    \frac{1}{C}OPT_{-i^*} -\frac{\varepsilon}{C}.
\end{displaymath}
Indeed, $L^*_{-i^*}\geq OPT_{-i^*}$ as $L$ is a fractional relaxation of $V$. Also, $OPT \leq OPT_{-i^*} + V(i^*)$,
hence,
\begin{equation}\label{eq:bound1}
    OPT\leq (1+C)V(i^*) + \varepsilon.
\end{equation}

If the condition  does not hold, by observing that $L^*_{-i^*}\leq L^*_c$ and
applying Proposition~\ref{prop:relaxation}, we get
\begin{displaymath}
    V(i^*)\leq \frac{1}{C}L^*_{-i^*} + \frac{\varepsilon}{C}
    \leq \frac{1}{C} \big(2 OPT + 2 V(i^*)\big) + \frac{\varepsilon}{C}.
\end{displaymath}
Applying Lemma~\ref{lemma:greedy-bound},
\begin{displaymath}
    V(i^*) \leq \frac{1}{C}\left(\frac{2e}{e-1}\big(3 V(S_G)
    + 2 V(i^*)\big) + 2 V(i^*)\right) + \frac{\varepsilon}{C}.
\end{displaymath}
Thus, if $C$ is such that $C(e-1) -6e  +2 > 0$,
\begin{align*}
    V(i^*) \leq \frac{6e}{C(e-1)- 6e + 2} V(S_G) 
    + \frac{(e-1)\varepsilon}{C(e-1)- 6e + 2}.
\end{align*}
Finally, using Lemma~\ref{lemma:greedy-bound} again, we get
\begin{equation}\label{eq:bound2}
    OPT(V, \mathcal{N}, B) \leq 
    \frac{3e}{e-1}\left( 1 + \frac{4e}{C (e-1) -6e  +2}\right) V(S_G)
    + \frac{2e\varepsilon}{C(e-1)- 6e + 2}.
\end{equation}
To minimize the coefficients of $V_{i^*}$  and $V(S_G)$  in \eqref{eq:bound1}
and \eqref{eq:bound2} respectively, we wish to chose $C$ that minimizes
\begin{displaymath}
    \max\left(1+C,\frac{3e}{e-1}\left( 1 + \frac{4e}{C (e-1) -6e  +2}
            \right)\right).
\end{displaymath}
This function has two minima, only one of those is such that $C(e-1) -6e
+2 \geq 0$. This minimum is
\begin{equation}\label{eq:constant}
    C =  \frac{8e-1 + \sqrt{64e^2-24e + 9}}{2(e-1)}.
\end{equation}
For this minimum, $\frac{2e\varepsilon}{C^*(e-1)- 6e + 2}\leq \varepsilon.$
Placing the expression of $C$ in \eqref{eq:bound1} and \eqref{eq:bound2}
gives the approximation ratio in \eqref{approxbound}, and concludes the proof
of Theorem~\ref{thm:main}.\hspace*{\stretch{1}}\qed

\section{Proof of Theorem \ref{thm:lowerbound}}

Suppose, for contradiction, that such a mechanism exists. Consider two
experiments with dimension $d=2$, such that $x_1 = e_1=[1 ,0]$, $x_2=e_2=[0,1]$
and $c_1=c_2=B/2+\epsilon$. Then, one of the two experiments, say, $x_1$, must
be in the set selected by the mechanism, otherwise the ratio is unbounded,
a contradiction. If $x_1$ lowers its value to $B/2-\epsilon$, by monotonicity
it remains in the solution; by  threshold payment, it is paid at least
$B/2+\epsilon$. So $x_2$ is not included in the solution by budget feasibility
and individual rationality: hence, the selected set attains a value $\log2$,
while the optimal value is $2\log 2$.\hspace*{\stretch{1}}\qed
