Suppose that we want to optimize the convex function $f(x)$ under some convex constraints
\begin{align}
&\underset{\mathbf{x}}{\operatorname{minimize}}& & f(\mathbf{x}) \\
&\operatorname{subject\ to}
& &g_i(\mathbf{x}) \leq 0, \quad i = 1, \dots, m \\
&&&h_i(\mathbf{x}) = 0, \quad i = 1, \dots, p
\end{align}
Assume that if we dispense with the constraints, the minimum of $f(x)$ does not satisfy the constraints (if we dispense with the constraints, the minimum of $f(x)$ does not lie on the feasible set).
Then, can we be sure that the solution of the constrained problem, $x^*$, lies on the apex of the feasible set? i.e:
\begin{align}
&& g_i(\mathbf{x^*}) = 0, \quad i = 1, \dots, m \\
&& h_i(\mathbf{x^*}) = 0, \quad i = 1, \dots, p
\end{align}
Can we be sure that the solution of the constrained problem, $x^*$, lies on the boundary of the feasible set? i.e:
\begin{align}
&& g_i(\mathbf{x^*}) = 0,\ for\ at\ least\ one \quad i = 1, \dots, m \\
&& h_i(\mathbf{x^*}) = 0, \quad i = 1, \dots, p
\end{align}
If the answer is no, is there any assumption or condition that we can add to this problem such that it guarantees that the solution of the constrained problem lies on the boundary or apex of the feasible set?
Best Answer
Not in general. Let's consider the in the standard setting, where we assume that $(g_i)_{i=1}^m$ and $(h_i)_{i=1}^p$ are convex. Even for this case, we are guaranteed to get a "sharp" constraint, i.e., at least one $g_i$ which meets their inequality $\leq$ with equality $=$. However, this does not mean that all of the constraints are sharp simultaneously!