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Let $X$ be a compact metric space. If $f:X\rightarrow \mathbb{R}$ is lower semi-continuous, then $f$ is bounded from below and attains its infimum. I want to prove this.

This is my proof:

Since $X$ is compact then it follows that $f(X)$ is compact therefore it is closed and bounded (because $f(K)$ is in $\mathbb{R}$) and therefore $f$ has an infimum, $m = \inf f(x)$. Let $(x_n)$ be a sequence in $X$. Since $X$ is compact by equivalence it is sequentially compact. There is a sub-sequence $(x_{n_k})$ of $(x_n)$ such that $f(x_{n_k})\rightarrow m \in\mathbb{R}$ since it is closed. Then by the lower semi-continuity of $f$ we have that $$m = \lim_{n\rightarrow\infty}\inf(x_n) = \lim_{n\rightarrow\infty}\inf(x_{n_k})\geq f(x) \geq m. $$ Thus $$m\geq f(x) \geq m.$$ Hence $m = f(x)$. Therefore $f$ attains its minimum and is bounded below.

(I did not assume $X$ was closed and bounded because it is not necessarily a subset of $\mathbb{R}^n$). This also is not homework I am trying to prove the theorems my book left blank to see if I really understand concepts. Any help and comments would be greatly appreciated. Thank you in advance.

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  • $\begingroup$ I don't think that $f(X)$ is necessarily closed. $\endgroup$ – triple_sec Sep 27 '13 at 6:14
  • $\begingroup$ @T.Bongers But $f$ is only lower semi-continuous. E.g., if $X=[0,1]$ and $$f(x)=\begin{cases}0&\text{if $x=0$,}\\x+1&\text{if $x\in(0,1],$}\end{cases}$$ then $f(X)=\{0\}\cup(1,2]$, which is not closed. $\endgroup$ – triple_sec Sep 27 '13 at 6:16
  • $\begingroup$ You have left out many steps, and produced an $x$ from nowhere in the first formula. I have given a standard line of reasoning below. $\endgroup$ – copper.hat Sep 27 '13 at 6:27
  • $\begingroup$ I thought that a subset of $\mathbb{R}^n$ is sequentially compact if and only if it is closed and bounded? $f(X)$ is a compact subset of $\mathbb{R}$. $\endgroup$ – RDizzl3 Sep 27 '13 at 6:56
  • $\begingroup$ Actually sorry I missed something in the theorem. It says if $X$ is compact and $f$ continuous then $f(X)$ is compact. That means it has to be both LSC and USC. $\endgroup$ – RDizzl3 Sep 27 '13 at 6:57
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Since $m = \inf f$, there exists a sequence $x_n$ such that $f(x_n) \to m$. This is true just by the definition of $\inf$.

Since $x_n \in X$ and $X$ is compact, we have some $\hat{x} \in X$ and some subsequence such that $x_{n_k} \to \hat{x}$. This follows since $X$ is compact, and hence sequentially compact (equivalent in a metric space in any case).

Since $f$ is lsc., we have $\liminf_n f(x_{n_k}) \ge f(\hat{x})$. This is by definition of lsc.

Since $\liminf_n f(x_{n_k}) = \lim_n f(x_{n_k}) = m$, we have $m \ge f(\hat{x})$, and by definition of $m$, we have $m \le f(\hat{x})$, hence $\hat{x}$ is a minimizer.

Note: The above proof does not depend on $m$ being finite (the proof shows that $m$ is finite, since $m \ge f(\hat{x})$). However, it is straightforward to show that $m$ is finite directly.

Since $f$ ls lsc., at any $x$, for all $\epsilon>0$, there is a $\delta >0$ such that if $y \in B(x,\delta)$, then $f(y) > f(x)-\epsilon$.

So, choose $\epsilon=1$, then for each $x$ we have some $\delta_x >0$ such that if $y \in B(x,\delta_x)$, then $f(y) > f(x)-1$. The $B(x,\delta_x)$ form an open cover of $X$, hence by compactness, there is a finite subcover $B(x_k,\delta_{x_k})$. Hence $f(x) > \min(f(x_1),...,f(x_n))-1$, from which it follows that $m \ge \min(f(x_1),...,f(x_n))-1$.

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  • $\begingroup$ Comments are not for extended discussion. I got the impression that you were done with the latest exchange, so this conversation has been moved to chat where you can still review it, if the need persists. $\endgroup$ – Jyrki Lahtonen May 16 '17 at 7:21

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