Suppose we have the quadratic minimization problem \begin{equation} \min_x \frac{1}{2} x^TPx + q^Tx +r \end{equation} We know that when $P$ is symmetric positive semi-definite, but the optimality condition $Px=-q$ does not have a solution, the quadratic problem above is unbounded below. (Boyd, page 458)
Now, the question is can unboundedness happen in least-squares minimization (which is clearly a quadratic minimization )? \begin{equation} \min_w \|Xw-y\|_2^2 \end{equation} What is the interpretation then? (I have never encountered an unbounded least squares.)
As user Gerry Myerson has pointed out in comments, the least squares is bounded below by zero, thus its not unbounded. To see this, it is enough to expand the squared 2-norm term \begin{align} \|Xw-y\|_2^2&=(Xw-y)^T(Xw-y) \\ &=w^T(X^TX)w-2y^TXw+y^Ty \\ &=w^T(X^TX)w-2(X^Ty)^Tw+y^Ty \end{align} Thus in this case, $P=X^TX$ and $q=X^Ty$ (neglecting the constant for the moment). Thus $q$ always lies in the range-space of $P$ and also $P$ is positive-definite or semi-definite (depending on rank of $X$).