Optimization with parametric constraints: solution maps

Question

For constrained optimization problems

$$ \begin{array}{ll} \min\limits_{x \in \mathbb R^n} & f(p, x) \\ \text{s.t.} & x \in C \end{array} $$

where $p \in \mathbb R$ is a parameter, we can deduce, under suitable convexity conditions, existence and Lipschitzness (in $p$) of solution maps $x^*(p) \in \arg\min\limits_{x \in C} f(p, x)$, say, by applying Theorem 2F.10 from this Rockafellar's book. Here, I attached an excerpt from there:

Question: can we extend this result to the case when $C$ depends on $p$? Say, it is described by $g(p, x) \le 0$ with whatever properties required, like continuity in $p, x$, convexity in $x$ for each $p$.

The said book does not seem to contain a result in such a generic form.

One particular problem I see there is that the subgradients be pivoted to the static constraint. The argument collapses when the constraint changes

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I'm almost sure that if the cost function is convex and the constraint doesn't change arbitrarily fast, the optimum must be continuous with a bounded rate of change

Answer 1

This is not a direct answer, but also too long for a comment.

You may be interested in Berge's Maximum Theorem, which I quote below from Aliprantis and Border (2006) (Theorem 17.31). A correspondence $F\colon X \twoheadrightarrow Y$ is a set-valued function that maps each $x\in X$ into a subset $Y$. I will define continuity for correspondences after the statement of the theorem.

Theorem (Berge). Let $\varphi\colon X\twoheadrightarrow Y$ be a continuous correspondence between topological spaces with nonempty compact values, and suppose $f\colon X\times Y\to\mathbb R$ is continuous. Define the "value function" $m\colon X \to \mathbb R$ by $$ m(x) = \max_{y\in\varphi(x)} f(x,y), $$ and the correpsondence of maximisers by $$ \mu(x) = \{ y \in \varphi(x) : f(x,y) = m(x) \}. $$ Then,

1. the value function is continuous;
2. the argmax correspondence $\mu$ is upper hemicontinuous and has nonempty compact values.

The statement of this theorem in Aliprantis and Border is actually slightly more general than this (they require only continuity of $f$ on the graph of $\varphi$), but I have taken liberties for simplicity.

A neighbourhood of a set $A$ is any set $B$ containing an open set $V$ such that $A \subset V \subset B$.

We say that a correspondence $\varphi\colon X \twoheadrightarrow Y$ is upper hemicontinuous (uhc) if for every $x \in X$, and every neighbourhood $U$ of $\varphi (x)$, there is a neighbourhood $V$ of $x$ such that $z \in V$ implies that $\varphi (z) \subset U$.

A correspondence $\varphi\colon X \twoheadrightarrow Y$ is lower hemicontinuous (lhc) if for every $x \in X$ and every open set $U$ such that $\varphi (x) \cap U \neq \emptyset$, there is a neighbourhood $V$ of $x$ such that $\varphi (z) \cap U \neq \emptyset$ whenever $z \in V$.

A correspondence is continuous if it is both uhc and lhc.

One might wonder whether it is possible to find a continuous function $g$ such that $g(x) \in \mu(x)$ for each $x$, so that $g$ is a continuous selection from $\mu$. However, as we can see above, we can only guarantee that $\mu$ is uhc, and continuous selection theorems I am familiar with usually require that the correspondence we are selecting from be lhc.

However, since we know that $\mu$ is uhc, we also know that $\mu$ is continuous when viewed as a function from $X$ into $Y$ whenever we can guarantee that $\mu(x)$ is a singleton for each $x$. A sufficient condition for this is for $f$ to be strictly concave in $y$ for each $x$, with $\varphi$ being convex-valued (that is, $\varphi (x)$ is a convex set for each $x$).

To see this, suppose that $y,y^\prime \in \mu(x)$, with $y\neq y^\prime$. Then, from strict concavity of $f(\cdot ,x)$ and the fact that $\varphi(x)$ is convex, we have that, for $\alpha \in (0,1)$, $$ f(x, \alpha y + (1-\alpha) y^\prime) > \alpha f(x,y) + (1-\alpha) f(x,y^\prime) = m(x)$$ and $\alpha y + (1-\alpha) y^\prime \in \varphi (x)$, a contradiction. This means that $\mu$ is uhc and singleton-valued, and is therefore continuous when viewed as a function.