Interpreting the dual of least norm solution of linear equations I'm trying to interpret the two results when deriving the dual of the convex optimization problem:
$$\text{minimize }\|x\|_2$$
$$\text{subject to }Ax=b$$
where we assume that $x \in \mathbb{R}$; that the domain, $D$, of the problem is the set of reals.
We begin by writing the problem's Lagrangian:
$$L(x, \nu)=\|x\|_2+\nu^TAx - \nu^Tb$$
and the dual function:
$$g(\nu)=\underset{x \in D}{\inf}L(x,\nu)=\underset{x \in D}{\inf}(\|x\|_2+\nu^TAx - \nu^Tb)$$

Solution #1: minimization by zero gradient condition
Standard practice at this point for differentiable $L(x,\nu)$ seems to be to find the optimal $x^*$ that satisfies:
$$\nabla_xL(x,\nu)=\frac{x}{\|x\|_2} + A^T\nu=0$$
which gives:
$$x^* = \left\{\begin{array}{ll}
-A^T\nu & \|A^T\nu\|_2=1 \\
-\infty & otherwise
\end{array}\right.$$
Then, substituting $x^*$ into $g(\nu)$ we have one (incomplete) interpretation of the dual function (for $ \|A^T\nu\|_2=1 $):
$$\begin{align}
g(\nu) &= \|-A^T\nu\|_2 - \|A^T\nu\|^2_2-b^T\nu\\
&= (1-1)-b^T\nu\\
&=
\left\{\begin{array}{ll}
-b^T\nu & \|A^T\nu\|_2=1\\
-\infty & otherwise
\end{array}\right.
\end{align}$$

Solution #2: general minimization over x
Alternately, we can choose to analyze the dual function to minimize the lagrangian over the domain of $x$ to give a more complete picture of the dual function. Again we start with the definition of the dual function:
$$g(\nu) = \underset{x\in D}{\inf}L(x,\nu)=\underset{x\in D}{\inf}(\|x\|_2+\nu^TAx - b^T\nu)
$$
and we make the choice of $x=-t\cdot A^T\nu, t\ge0$ (though $x=t\cdot A^T\nu, t\le0$ should work similarly) and substitute into $g(\nu)$ giving:
$$
\begin{align}
g(\nu) &= \underset{t\ge0}{\inf}(t\cdot(\|A^T\nu\|_2 - \|A^T\nu\|_2^2) - b^T\nu) \\
&= \left\{\begin{array}{ll}
-b^T\nu & \|A^T\nu\|_2 \le 1 \\
-\infty & otherwise
\end{array}\right.
\end{align}
$$

Question:
In solution #2 we get the additional finite expression, $-b^T\nu$, for the dual function when $\|A^T\nu\|_2\lt 1$.
My questions are:


*

*what is the interpretation of the different solutions when $\|A^T\nu\|_2\lt1$?   

*why is it necessary to restrict $t$ to be non-negative in the substitution step for solution #2? How can we justify this constraint, and does this constraint influence the answer to question #1?


I'm looking for some geometric intuition into this problem that can explain the apparent difference here. I appreciate the help!
 A: This is an elaboration of my comment. 
Michael's suggestion hits the nail on the head as it avoids the pitfalls encountered
in both Solution #1 & Solution #2.
In both cases, the issue is that the objective is not differentiable everywhere, and so a little care is needed in the characterisation of a solution. To get some intuition, plot the function $x \mapsto |x|+bx$ for $|b|<1, |b|=1, |b|>1$ to see how the infimal value
varies with $b$.
Regarding Question #1, let $\lambda(x) = L(x,\nu)$, we are looking for a solution of $\inf_x \lambda(x)$. Note that $\lambda$ is not
differentiable everywhere, so we need to take care at points where
it is not differentiable. Since it is convex and defined everywhere,
we can use the subdifferential instead.
Since $\lambda$ is convex, we have
a minimiser (that is, some $\hat{x}$ such that $\lambda(\hat{x})=\inf_x \lambda(x)$) iff $0 \in \partial \lambda (x)$ for some $x$, where $\partial \lambda (x)$ is the subdifferential at $x$.
We have $\partial \lambda (x) = \partial \|\cdot\|_2 (x) + \{ A^T \nu \} = \begin{cases} \overline{B}(0,1)+ \{ A^T \nu \}, & x=0 \\ \{ { x \over \|x\|_2 } \} + \{ A^T \nu \}, & \text{otherwise}\end{cases}$.
From this we see that there is no minimiser if $\|A^T \nu \|_2 > 1$, the minimiser is $\hat{x}=- A^T \nu$ if $\|A^T \nu \|_2 = 1$, and
the minimiser is $\hat{x}=0$ if $\|A^T \nu \|_2 < 1$.
From this we get $\inf_x \lambda(x) = \begin{cases} -\nu^T b, & \|A^T \nu \|_2 \le 1\\ -\infty, & \text{otherwise}\end{cases}$, which
matches Solution #2. (Note that the non existence of a minimiser
does not imply that the function is unbounded below, but it is easy 
to explicitly choose an $x$ in this case that shows that the function is
unbounded below.)
Regarding Question #2, as you have (mostly) observed, we have
$\inf_x \lambda(x) = \inf_t \lambda(t A^T \nu) $. Expanding,
$\lambda(t A^T \nu) = |t| \|A^T \nu\|_2 - t \|A^T \nu \|_2^2 - \nu^T b$ (note again the non differentiability at $t=0$ because of the $|t|$ term).
We can either perform a case analysis or compute the subdifferential
as above to compute 
$\inf_t \lambda(t A^T \nu) = \begin{cases} -\nu^T b, & \|A^T \nu \|_2 \le 1\\ -\infty, & \text{otherwise}\end{cases}$.
