Let $X$ denote the non-negative "orthant" of the Banach space $L^2$ (or whatever you call the set of functions in $L^2$ that are non-negative), and let $C$ be a closed, convex subset of $X$. Let $f$ be a strictly convex functional defined on $X$ and suppose that the problem $$\text{minimize}~ f(x)~~~s.t.~~~x\in C$$admits a unique solution $x_0\in C$. Further, suppose that there is also a unique "tangent plane" to $C$ at $x_0$, i.e., a unique functional $y_0 \in L^2$ such that $\left\langle x_{0},y_{0}\right\rangle = q$ and $\left\langle x,y_{0}\right\rangle\leq q$ for all other $x\in C$. Is it also true that $x_0$ minimizes the relaxed problem $$\text{minimize}~ f(x)~~~s.t.~~~\left\langle x,y_{0}\right\rangle\leq q~?$$ Could I make my problem more tractable by considering a different space of functions instead of $L^2$, e.g. continuous functions or $L^\infty$, for example?
1 Answer
Let us assume that your $f$ is at least directionally differentiable at $x_0$ and that the directional derivative depends continuously on the direction (this may be satisfied under rather general assumptions, e.g. if $f$ is locally Lipschitz).
Then, the first order (necessary and - due to convexity - sufficient) optimality conditions of your first problem read $$f'(x_0;d) \ge 0 \quad\forall d \in T_C(x_0),$$ where $T_C(x_0)$ is the tangent cone of $C$ in $x_0$. Since your functional $y_0$ is unique, we find $T_C(x_0) = y_0^\circ$, where $y_0^\circ = \{d : \langle d, y_0\rangle \le 0\}$ is the polar cone of $y_0$. This yields the optimality conditions $$f'(x_0;d) \ge 0 \quad\forall d \in y_0^\circ,$$ but these are precisely the first order (nec. and suff.) conditions of the second problem.
