On the Lower Semicontinuity of Parametric Optimization

When studying optimization problems with parameter-dependent feasible sets, a key technical step is showing that the value function behaves nicely — for example, that it is lower semicontinuous. In this post, we’ll prove such a property for a simple but fundamental setting.

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Setting

Let’s assume the following:

• $X \subset \mathbb{R}^n$ is compact.

• $f, g : X \to \mathbb{R}$ are continuous functions.

• For each real number $a$, define the feasible set

  $$S(a) := \{x \in X : g(x) \le a \}.$$

• Define the value function

  $$F(a) := \min_{x \in S(a)} f(x).$$

We first verify that the minimum value defining $F(a)$ is well-defined. Indeed, since $X$ is compact, it is also closed. The feasible set
$$S(a) = X \cap \{\, x \in X : g(x) \le a \}$$
is the intersection of two closed sets, and hence itself closed. As a closed subset of the compact set $X$, $S(a)$ is compact. Therefore, $F(a)$ represents the minimum of a continuous function over a compact set, and the minimum is attained whenever $S(a) \ne \emptyset$.

In the following, our goal is to show that $F$ is lower semicontinuous.
Because $F$ will turn out to be nonincreasing, lower semicontinuity is equivalent to right-continuity at every feasible $a_0$.

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Step 1. $F$ is nonincreasing

If $a_1 < a_2$, then $S(a_1) \subset S(a_2)$.
Therefore,

$$F(a_2) \le F(a_1),$$

so $F$ is nonincreasing.

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Step 2. Right continuity at $a_0$

Assume $S(a_0) \ne \emptyset$.
Take any sequence $a_n \downarrow a_0$ (that is, $a_n \ge a_0$ and $a_n \to a_0$).

Since $S(a_n) \supset S(a_0)$ for all $n$, and each $S(a_n)$ is compact, there exists $x_n \in S(a_n)$ such that

$$F(a_n) = f(x_n), \quad g(x_n) \le a_n.$$

Because $X$ is compact, the sequence $(x_n)$ has a convergent subsequence $x_{n_k} \to x^\ast \in X$.
By continuity of $g$,

$$g(x^\ast) = \lim_{k} g(x_{n_k}) \le \lim_{k} a_{n_k} = a_0,$$

so $x^\ast \in S(a_0)$.

By continuity of $f$,

$$\lim_{k} F(a_{n_k}) = \lim_{k} f(x_{n_k}) = f(x^\ast) \ge \min_{x \in S(a_0)} f(x) = F(a_0).$$

Since this holds for every subsequence, we have

$$\liminf_{n \to +\infty} F(a_n) \ge F(a_0).$$

Because the sequence $a_n \downarrow a_0$ was arbitrary, we obtain

$$\liminf_{a \downarrow a_0} F(a) \ge F(a_0),$$

which means $F$ is right-continuous at $a_0$.

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Conclusion

Since $F$ is nonincreasing, right-continuity is equivalent to lower semicontinuity. Therefore, for all $a_0$ such that $S(a_0) \ne \emptyset$,

$$F(a) = \min{, f(x) : g(x) \le a, , x \in X ,}$$

is lower semicontinuous.

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Further Notes

• The same reasoning still works if $f$ is only lower semicontinuous, by using $\liminf f(x_n) \ge f(x^\ast)$ instead of full continuity.

• For upper semicontinuity or full continuity of $F$, one can invoke Berge’s Maximum Theorem, which provides conditions involving closed graphs and semicontinuity of the feasible set mapping.