This post will focus on the elementary derivation of taking the Lagrangian dual of the LASSO program. In full transparency, I must admit, having taken convex optimization a while ago, I forgot the details on how to do this off the top of my head. Plus, I am TA-ing for Berkeley's undergraduate machine learning course this upcoming semester, so I figured it was worthwhile to refresh myself on the basics. $\newcommand{\abs}[1]{| #1 |} \newcommand{\ind}{\mathbf{1}} \newcommand{\norm}[1]{\lVert #1 \rVert} \newcommand{\R}{\mathbb{R}} \newcommand{\ip}[2]{\langle #1, #2 \rangle} \newcommand{\T}{\mathsf{T}} \newcommand{\Proj}{\Pi} \newcommand{\Tr}{\mathrm{Tr}}$

Setup: Fix a linear operator $A \in \R^{m \times n}$, response vector $b \in \R^{m}$, regularization parameter $\lambda > 0$, and integer power $p \in \{1, 2\}$. The LASSO program is $$P^* := \min_{x \in \R^n} \;\; \norm{Ax - b}^p_2 + \lambda \norm{x}_1 \:.$$ It is well known that the $\ell_1$-norm is a sparsity-inducing norm, but I won't go into those details here. Let's work out the dual to this program instead.

Deriving the dual: To make our lives easier, we start by introducing the variable $\R^{m} \ni z = Ax$. The tautologically equivalent program to the LASSO, with this variable substitution, is $$P^* = \min_{x \in \R^n, z \in \R^m} \;\; \norm{z - b}^p_2 + \lambda \norm{x}_1 : z = Ax \:.$$ We now make the program unconstrained via another tautological change, by introducing the Lagrange multiplier $\Lambda \in \R^m$, $$$$P^* = \min_{x \in \R^n, z \in \R^m} \max_{\Lambda \in \R^m} \;\; \norm{z - b}^p_2 + \lambda \norm{x}_1 + \ip{\Lambda}{Ax - z} \:. \label{eq:starting_point}$$$$ Equation $\eqref{eq:starting_point}$ is our starting point for analysis. By Slater's condition, we can swap the min and max in $\eqref{eq:starting_point}$ to conclude \begin{align*} P^* &= \max_{\Lambda \in \R^m} \min_{x \in \R^n, z \in \R^m} \;\; \norm{z - b}^p_2 + \lambda \norm{x}_1 + \ip{\Lambda}{Ax - z} \\ &= \max_{\Lambda \in \R^m} \left[ \min_{z \in \R^m} \;\; \left\{\norm{z - b}^p_2 - \ip{\Lambda}{z}\right\} + \min_{x \in \R^n} \;\; \left\{\lambda \norm{x}_1 + \ip{A^\T \Lambda}{x}\right\} \right] \\ &:= \max_{\Lambda \in \R^m} \;\; T_1( \Lambda ) + T_2( \Lambda ) \:. \end{align*} It is now evident why we made the substitution $z = Ax$, since we now have two separate problems $T_1(\Lambda)$ and $T_2(\Lambda)$, to work on, each of which individually are easy.

Let us work on $T_2(\Lambda)$ first, since it is independent of our choice of $p$. We see that it is actually a coordinate separable program, so we work on each coordinate at a time. For every $i \in \{1, ..., n\}$, we need to solve $$T_{2,i}(\Lambda) := \min_{x_i \in \R} \;\; \lambda \abs{x_i} + (A^\T \Lambda)_i x_i = \min_{x_i \in \R} \;\; (\lambda + \mathrm{sign}(x_i) (A^\T \Lambda)_i ) \abs{x_i} \:,$$ where for a vector $v$, $(v)_i$ denotes the $i$-th coordinate. But it is easy to compute the value of $T_{2_i}(\Lambda)$ by considering the cases $$T_{2,i}(\Lambda) = \begin{cases} 0 &\text{if } \abs{(A^\T \Lambda)_i} \leq \lambda \\ -\infty &\text{o.w.} \end{cases} \:.$$ Hence, $$T_2(\Lambda) = \sum_{i=1}^{n} T_{2,i}(\Lambda) = \begin{cases} 0 &\text{if } \norm{A^\T \Lambda}_\infty \leq \lambda \\ -\infty &\text{o.w.} \end{cases} \:.$$

We now work on $T_1(\Lambda)$. If $p=2$, this is easy. Almost by inspection, we have that $$T_1(\Lambda) = - \left(\frac{1}{4} \norm{\Lambda}^2_2 + \ip{\Lambda}{b}\right) \:.$$ If $p = 1$, we need to do a little bit more work. We first perform some manipulations \begin{align*} T_1(\Lambda) = \min_{z \in \R^m} \;\; \norm{z - b}_2 - \ip{\Lambda}{z} = \min_{z \in \R^m} \;\; \norm{z}_2 - \ip{\Lambda}{z + b} = -\ip{\Lambda}{b} + \min_{z \in \R^m} \;\; \left\{ \norm{z}_2 - \ip{\Lambda}{z} \right\} \:. \end{align*} We now observe that $$\min_{z \in \R^m : z \neq 0} \;\; \norm{z}_2 - \ip{\Lambda}{z} = \min_{z \in \R^m : z \neq 0} \;\; \left(1 - \ip{\Lambda}{\frac{z}{\norm{z}_2}}\right) \norm{z}_2 = \min_{\alpha > 0} \;\; \left(1 - \norm{\Lambda}_2\right) \alpha \norm{\Lambda}_2 \:.$$ The last equality follows from the Cauchy-Schwarz inequality. In this form, it is now easy to see that $$\min_{\alpha > 0} \;\; \left(1 - \norm{\Lambda}_2\right) \alpha \norm{\Lambda}_2 = \begin{cases} 0 &\text{if } \norm{\Lambda}_2 \leq 1 \\ -\infty &\text{o.w.} \end{cases} \:.$$ Putting these calculations together, $$T_1(\Lambda) = \begin{cases} -\ip{\Lambda}{b} &\text{if } \norm{\Lambda}_2 \leq 1 \\ -\infty &\text{o.w.} \end{cases} \:.$$

We are now ready to put the final pieces together. If $p=2$, we have $$P^* = \max_{\Lambda \in \R^m} \;\; -\frac{1}{4} \norm{\Lambda}^2_2 - \ip{\Lambda}{b} : \norm{A^\T \Lambda}_\infty \leq \lambda \:.$$ On the other hand, if $p=1$, we have $$P^* = \max_{\Lambda \in \R^m} \;\; - \ip{\Lambda}{b} : \norm{\Lambda}_2 \leq 1, \norm{A^\T \Lambda}_\infty \leq \lambda \:.$$

The regime $m \ll n$. Observe that while the primal problem is an unconstrained optimization program over $\R^n$, the dual is a constrained program over $\R^m$. In the case when $m \ll n$, as is typically the case with LASSO (the system $Ax=b$ is very underdetermined), it can be computationally advantageous to work in the dual. However, there is no free lunch, as the dual introduces hard constraints.