path: root/dual2.tex

\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{amsmath, amsfonts, amsthm}
\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\input{definitions}

\begin{document}
Let $c$ be a cost vector in $[0,1]^n$, and $x_1,\ldots,x_n$, $n$ vectors in
$\mathbf{R}^d$ such that for all $i\in\{1,\ldots,n\}$, $b\leq \T{x_i}{x_i}\leq
1$ for some $b\in(0,1]$. Let us consider the following convex optimization
problem:
\begin{equation}\tag{$P_c$}\label{eq:primal}
    \begin{split}
    \mathrm{maximize}\quad & L(\lambda) \defeq\log\det\left(I_d + \sum_{i=1}^n
    \lambda_i x_i x_i^T\right)\\
    \textrm{subject to}\quad & c^T\lambda \leq 1 \;;\; 0\leq\lambda\leq \mathbf{1}
\end{split}
\end{equation}
We denote by $L^*_c$ its optimal value.

Let $\alpha\in\mathbf{R}^+$, consider the perturbed optimization problem:
\begin{equation}\tag{$P_{c, \alpha}$}\label{eq:perturbed-primal}
    \begin{split}
        \mathrm{maximize}\quad & L(\lambda) \defeq\log\det\left(I_d + \sum_{i=1}^n
    \lambda_i x_i x_i^T\right)\\
    \textrm{subject to}\quad & c^T\lambda \leq 1 \;;\; \alpha\leq\lambda\leq \mathbf{1}
\end{split}
\end{equation}
and denote by $L^*_c(\alpha)$ its optimal value. Note that we have $L^*_c = L^*_c(0)$.

We will assume that $\alpha<\frac{1}{n}$ so that \eqref{eq:perturbed-primal} has at
least one feasible point: $(\frac{1}{n},\ldots,\frac{1}{n})$.

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: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}^n 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}

\begin{lemma}\label{lemma:proximity}
We have:
\begin{displaymath}
    L^*_c - \alpha n^2(n-1)\leq L^*_c(\alpha) \leq L^*_c
\end{displaymath}
In particular, $|L^*_c - L^*_c(\alpha)| \leq \alpha n^2(n-1)$.
\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 = \{i|\lambda_i^* = \alpha\}
    \end{displaymath}

    Let us first assume that, $|M|\geq 1$, then we have that $\T{c}\lambda^*
    = 1$, 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 (n-1) + (n-1)\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 - (n-1)\alpha}{n-1}> \frac{1}{n(n-1)}
    \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^* = \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 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(n-1)
    \end{displaymath}
    
    Finally let us write:
    \begin{displaymath}
        \T{\mathbf{1}}\mu^* = \sum_{i=1}^n \mu^*_i = \sum_{i\in M}\mu_i^*
    \end{displaymath}

    Either $|M|=0$, in which case $\T{\mathbf{1}}\mu^* = 0$, either $|M|\geq
    1$, in which case $\T{\mathbf{1}}\mu^*\leq n^2(n-1)$ from the inequality above.
    In both cases, $\T{\mathbf{1}}\mu^* \leq n^2(n-1)$, 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, $\mathcal{L}_{c', \alpha}(\lambda, \mu^*, \nu^*, \xi^*) \geq
    \mathcal{L}_{c, \alpha}(\lambda, \mu^*, \nu^*, \xi^*) + \lambda_i\xi^*\delta$, 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}

\begin{proposition}
    Let $\delta\in(0,1]$. For any $\varepsilon\in(0,1]$, there exists
    a routine 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}

\begin{proof}
Let $\varepsilon$ in $(0, 1]$. The routine works as follows: set $\alpha\defeq
\varepsilon(\delta + n^2(n-1))^{-1}$ 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 requested.

\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(n-1) = \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(n-1))}
\end{displaymath}
\sloppy
hence, the standard convex optimization algorithm runs in time $O(poly(n, d,\log\log A^{-1}))$. Note that:
\begin{displaymath}
    \log\log A^{-1} = O\bigg(\log n\; \log\log\frac{1}{\epsilon\delta b}\bigg)
\end{displaymath}
which yields the wanted running time for the routine.\qedhere
\end{enumerate}
\end{proof}
\end{document}