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.


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


  • 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 $J(x_n)\overset{n\to \infty}{\rightarrow} \min_{x \in C} J(x)$?


2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$
    – Hermi
    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 $$\lim_{n\to\infty} J(x_n)=\ell.$$

See also Lemma 2 in here.


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