Uniqueness of the solution to a quadratic problem [closed]

Question

Consider a positive definite matrix $\boldsymbol H$, the known vectors ${\boldsymbol b}$ and ${\boldsymbol a}_i$. Now the minimization problem is casted with respect to the vector ${\boldsymbol x} $ and the scaler $y$ : \begin{equation} \begin{split} &\min_{\boldsymbol x, y} \, {\boldsymbol x}^\top {\boldsymbol H} {\boldsymbol x} + {\boldsymbol b}^\top {\boldsymbol x} + y \\ & \text{subject to } -{\boldsymbol a_i^\top \, x} + y \geq 0, \; \text{ for }\forall i \end{split} \end{equation} Although the Hessian of the objective function $\left(\begin{array}{cc}{\boldsymbol H} & {\boldsymbol 0}\\{\boldsymbol 0}^\top &0 \end{array}\right) $with respect to $[{\boldsymbol x}^\top, y]^\top$ is semi-definite positive, I think the problem admits an unique solution, since we can replace $y$ in the objective by $\max_i {\boldsymbol a}_i^\top {\boldsymbol x}$, which is convex.

If this is correct, could someone provides formal explanations using convex optimization theory to support this uniqueness? Thank you.

Answer 1

If you replace $y$ as you suggest, then your objective (only in terms of $x$) is given by a strongly convex function (the quadratic) plus a convex function. Since this defines a strongly convex objective you have uniqueness guaranteed. Note however that uniqueness for $x$ does not immediately imply uniqueness of $(x,y)$, but probably you are not concerned much about $y$.

I hope this helps.