Commit ca1c63c9 authored by nanored's avatar nanored

début dm 3

parent 86541fb0
\documentclass[11pt]{article}
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\usepackage{amsmath,amssymb,amsfonts}
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\title{
\noindent\rule{\linewidth}{0.4pt}
\huge Convex Optimization --- DM 3
\noindent\rule{\linewidth}{1pt}
}
\author{Yoann Coudert-\,-Osmont}
\newcommand{\trans}{\mathsf{T}}
\newcommand{\syst}[2]{\left\{ \begin{array}{#1} #2 \end{array} \right.}
\newcommand{\diag}{\text{\textbf{diag}}~}
\newcommand{\pmat}[1]{\begin{pmatrix} #1 \end{pmatrix}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\one}{\mathbbm{1}}
\begin{document}
\maketitle
\section*{Question 1}
We transform our primal by adding a new variable in order to add a constraint:
$$ \begin{array}{llr}
\min_{w, z} & \dfrac{1}{2} \| z \|_2^2 + \lambda \| w \|_1 \\[2mm]
\text{s.t.} & X w - y = z
\end{array} $$
Then, we compute the dual function of this new formulation:
$$ \begin{array}{lll}
g(v) & = & \inf_{z, w} \dfrac{1}{2} \| z \|_2^2 + \lambda \| w \|_1 + v^\trans \left( Xw - y - z \right) \\[2mm]
& = & - v^\trans y + \inf_{z, w} - v^\trans z + \dfrac{1}{2} \| z \|_2^2 + \left( X^\trans v \right)^\trans w + \lambda \| w \|_1 \\[3mm]
& = & - v^\trans y - \dfrac{1}{2} \| 2 v \|_2^{2 \, *} - \lambda \left\| - \dfrac{1}{\lambda} X^\trans v \right\|_1^*
\end{array} $$
In the previous homework I compute the conjugate of the square of 2-norm and the conjugate of the 1-norm. We have :
$$ \| x \|_2^{2 \, *} = \frac{1}{4} \| x \|_2^2 \qquad \| x \|_1^* = \syst{ll}{
0 & \text{If } \| x \|_{\infty} \leqslant 1 \\
+ \infty & \text{Otherwise}
} $$
We then simplify the dual function:
$$ g(v) = \syst{ll}{
- \dfrac{1}{2} \| v \|_2^2 - y^\trans v & \text{If } \left\| X^\trans v \right\|_\infty \leqslant \lambda \\
- \infty & \text{Otherwise}
} $$
Furthermore we can rewrite the condition by :
$$ \forall i = 1, ..., d \; \left| \left( X^\trans v \right)_i \right| \leqslant \lambda \Leftrightarrow X^\trans v \preceq \lambda \one_d \text{ and } - X^\trans v \preceq \lambda \one_d $$
Then we let the matrix:
$$ A = \pmat{X^\trans \\ -X^\trans} \qquad \text{such that } A v \preceq \lambda \one_{2d} $$
We also remark that the following equality:
$$ \dfrac{1}{2} \| v \|_2^2 = \dfrac{1}{2} v^\trans I_n v = v^\trans Q v \qquad \text{with } Q = \dfrac{1}{2} I_n $$
The dual problem is obtained by maximizing $g$ which is equivalent to minimizing $-g$. Thus we obtain the following dual:
$$ (QP) \qquad \begin{array}{llr}
\min_v & v^\trans Q v + y^\trans v \\[2mm]
\text{s.t.} & A v \preceq \lambda \one_{2d}
\end{array} $$
We identify with the notations of the statement :
\begin{center}
\fbox{$ \displaystyle Q = \dfrac{1}{2} I_n \quad p = y \quad A = \pmat{X^\trans \\ -X^\trans} \quad b = \lambda \one_{2d} $}
\end{center}
\end{document}\texttt{}
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