TITLE: Removing constraints in convex optimization
QUESTION [9 upvotes]: Say I have a convex optimization problem of the form $$\min_x f(x) ~~ s.t.\\ g_1(x)\leq0,\\\vdots \\g_n(x)\leq 0$$ with all functions convex. Suppose that $x^*$ is a unique optimizer to my problem and that we have $g_i(x^*)=0$ for some index $i$ and that $g_j(x^*)<0$ for all other indices.  Then, it is well-understood that $x^*$ is also a solution to the problem $$\min_x f(x) ~~ s.t.\\ g_i(x)\leq0~.$$ in other words, the $(n-1)$ inactive constraints did not contribute anything to the problem.  My question is:  does anything change if, instead of having $n$ constraint functions, I have a continuum of functions?  That is, say that I now have the problem $$\min_x f(x) ~~ s.t.\\ g_t(x)\leq0 ~ \forall t\in[0,1] ~ $$ for some parameterized family of functions $g_t$, and I can again show that there is a unique minimizer $x^*$ where there is a single value of $t^*$ such that $g_{t^*}(x^*)=0$ and $g_{t}(x^*)<0$ otherwise.  Are there any extra precautions I need to take before claiming that it will suffice to consider the simpler problem $$\min_x f(x) ~~ s.t.\\ g_{t^*}(x)\leq0 ~ $$?  My gut tells me no, but strange things are known to happen at infinity.

REPLY [4 votes]: Part of the reason this works for a finite set of constraints, and in particular
the reason that the simplified problem still has a unique minimizer, is that
$x^*$ is an interior point of each of the regions $g_j(x) \lt 0,\; j \ne i$ and
therefore is an interior point of their intersection. This does not work if 
there are infinitely many constraints, and you cannot rely on uniqueness.
For a simple example consider minimizing $x_2$ under the constraint 
$x_1^2 - x_2 \le 0$. Clearly the unique minimizer is $(0, 0)$. We can however replace the single strictly convex constraint by an equivalent family of
linear constraints $g_t(x) = 2tx_1 - x_2 - t^2 \le 0,\; t \in \mathbb{R}$. 
The only active constraint is then $g_0(x) \le 0$, but if we remove all the other constraints, the (over)simplified problem has every
$(x_1, 0),\; x_1 \in \mathbb{R}$ as a minimizer.