# Theorem of Weierstrass for Closed Functions

In a lecture, we stated and proved the following theorem of Weierstrass, but I would have a question to it. First, let me give some basic definitions and assumptions before stating it.

Assumptions:

• $$X$$ is a finite vector space.
• Let $$J:X\rightarrow\mathbb R_{\infty}:=\mathbb R\ \cup \ \{ \infty \}$$ be an extended real function.

Definitions:

• The effective domain of $$J$$ is the set of points where $$J$$ is finite, i.e. $$\text{dom}(J) := \{ x\in X \ \vert \ J(x) < \infty \}.$$
• The function $$J: X\mathbb \rightarrow R_{\infty}$$ is said to be proper if $$\text{dom}(J) \ne \emptyset$$.
• A function $$J: X\rightarrow \mathbb R_{\infty}$$ is closed if its epigraph, $$\text{epi}(J) := \{ (x, \alpha) \in X\times \mathbb R \ \vert \ J(x) \leq\alpha \},$$ is closed. This is equivalent (by a Theorem) to $$J$$ being lower semi-continuous, i.e. for any sequence $$\{x_n\}_n\subset X$$ with $$x_n\rightarrow x$$ it holds that $$J(x^{\star}) \leq \lim\inf_{n\to\infty}J(x_n)$$.

Now we proved: Theorem: Let $$J:X\rightarrow\mathbb R_{\infty}$$ be proper, closed and let $$C\subset X$$ be compact with $$C \ \cap \ \text{dom}(J) \ne \emptyset$$. Then $$J$$ attains its minimal value over $$C$$. Since $$J$$ is closed ($$\Leftrightarrow J$$ is lower semi-continuous) $$\Rightarrow J(x^{\star}) \leq \lim\inf_{k\to\infty}J(x_{n_k})$$

Proof (in our lecture): There exists a sequence $$\{x_n\}_{n}$$ with $$J(x_n)\overset{n\to \infty}{\rightarrow} \min_{x \in C} J(x)$$. According to the Bolzano-Weierstrass theorem, there exists a convergent subsequence $$\{x_{n_k}\}_{k}\overset{k\to\infty}{\rightarrow} x^{*}\in C$$. Since $$J$$ is lower semi-continuous $$\Rightarrow J(x^{\star}) \leq \lim\inf_{k\to\infty} J(x_{n_k}) = \lim_{k\to\infty}J(x_{n_k}) = \min_{x\in C}J(x)$$. Thus, $$x^{\star} \in C$$ is the minimizer. QED

Question: Why is it possible to choose a sequence $$\{x_n\}_{n}$$ with $$J(x_n)\overset{n\to \infty}{\rightarrow} \min_{x \in C} J(x)$$?

You need to make sure that you are choosing the $$x_n$$ such that $$\lim_{n\rightarrow\infty}J(x_n)=\inf\{J(C)\}.$$
Then by the lower semi-continuity you get that $$J(x^*)\leq\lim\inf J(x_{n_k})$$, but by the above equality you also get that $$\forall \epsilon > 0$$, eventually $$J(x_{n_k})\leq \inf\{J(C)\} +\epsilon\leq J(x^*) + \epsilon$$.
Thus $$J(x^*)\geq \lim\inf J(x_{n_k})$$ and you have equality.
• I apologize, I realize I made a mistake. I'll update my post. My question would now be: Why is it possible to choose a sequence $\{x_n\}_{n}$ with $J(x_n) \overset{n\to \infty}{\rightarrow} \min_{x\in C}J(x)$? You did sth very similar with the infimum, but why can we do this? Commented Jun 10, 2021 at 11:55
Let $$\ell:=\inf_{x\in C}J(x)$$ for simplicity. If $$\ell=J(x^*)$$ for some $$x^*\in C$$, then choose $$x_n=x^*$$ for every $$n\geq1$$. Otherwise, by the definition of $$inf$$, for every $$n\geq1$$ there is $$x_n\in C\cap \mathrm{dom}(J)$$ such that $$J(x_n)\in [\ell,\ell+1/n]$$. Hence $$\lim_{n\to\infty} J(x_n)=\ell.$$