# Is this nonnegativity constraint convex?

Clearly, a Linear quadratic regulator problem involves quadratic cost which is convex. And minimizations that involves convex objective functions have unique solution...or global minimum. I understand that in general if the objective function is convex and the constraints are also convex, then the task of deriving optimal solution remains in the area of convex optimization. But, what if the LQR problem is constrained? such as the states are constrained to be nonnegative $x(t)\ge 0$? What can we say about the convexity of this constraint?

• A convex function has a unique minimum (if it has one: this may be not attained and actually be an infimum), but not a unique minimizer in general. Additional assumptions may be needed to have that. – Lorenzo Stella Jul 31 '16 at 13:23
• @LorenzoStella Can you mention those essential assumptions for the existence of unique minimizer. The problem assumes that the system is controllable. Thanks – user252783 Aug 6 '16 at 7:05
• For example, strong convexity implies existence of a unique minimum point, but I guess that is rarely the case for this type of problems, right? – Lorenzo Stella Aug 6 '16 at 9:47

Long answer: the problem is to understand what are the decision variables of your problem, with respect to which you classify it as convex or nonconvex. The initial state $x_0 \in \mathbb{R}^{n_x}$ is usually considered as a parameter, and the inputs $u_0,u_1,\ldots,u_{N-1} \in \mathbb{R}^{n_u}$ as decision variables, as well as subsequent states in time, $x_1,\ldots,x_N \in \mathbb{R}^{n_x}$. States and inputs are bound to each other by the dynamics, which is a constraint in this setting: given that constraints on the $x$'s and the $u$'s are convex (like nonnegativity of the states, as you mentioned), the problem is certainly convex if the dynamics is linear, i.e. $$x^{k+1} = Ax^k+Bu^k,\quad k=0,1,\ldots,N-1$$ where $A$ and $B$ are linear maps (matrices). If instead the dynamics non nonlinear, the problem is nonconvex in general. This is simply because in this case the feasible set (with respect to the linear dynamics above) is an affine subspace, which is convex in particular. Intersecting this with convex constraints on the $x$'s and the $u$'s still yields a convex feasible set.