Suppose that the least squares problem
$$\min_x \| Ax-b \|$$
has a unique solution $x^*$. Now, when we consider the constrained version such as
${\rm min}_x$ $\parallel Ax-b\parallel$ s.t. $x_i \leq 0$, $i=1,2,\dotsc,n$ (every $n$ element in vector $x$) can we say, without loss of generality that, there exists a solution for the constrained problem, only if the solution of the unconstrained problem $x^*$ lies inside the constraint set ? Or am I missing something?
My question is, if the unconstrained problem has a unique (one and only one possible) solution $x^*$ and if the feasible set derived with respect to the newly added constraint ($x_i \leq 0$, $i=1,2,\dotsc,n$ ) to the problem, does not include the "only possible =unique solution" of the unconstrained problem (${\rm min}_x$ $\parallel Ax-b\parallel$), then doesnt it mean that there does not exist a solution for the constrained problem?