extended real valued function bounded below and attains its minimal value

**morito14** · September 11th 2010, 01:56 PM

Let f(x) be a real-valued function (an extended real valued function) on a metric space (M; d). Then
$\lim\sup_{x\rightarrow y}f(x)=\lim_{\epsilon\rightarrow0}\sup_{0<d(x,y)<\ epsilon}f(x)$
and
$\lim\inf_{x\rightarrow y}f(x)=\lim_{\epsilon\rightarrow0}\inf_{0<d(x,y)<\ epsilon}f(x)$

A function is called lower (upper) semicontinuous at a point y if $f(y)\neq\pm\infty$
and
$f(y)=\leq\liminf_{x\rightarrow y}f(x)$ and
$f(y)=\geq\limsup_{x\rightarrow y}f(x)$

Let a metric space M be complete, and let K be a compact subset in M. Prove that a lower semicontinuous function on K is bounded from below and attains its minimal value and an upper semicontinuous function on K is bounded from above and attains its maximal value.

Any idea of how to approach this??

**Jose27** · September 11th 2010, 09:08 PM

Dealing with the first one (lower semicontinuity), assume f is not bounded below, then there exists a sequence $(x_n) \subset K$ such that $f(x_n) \rightarrow -\infty$ , but $K$ is compact, so it's sequentially compact, so without loss of generality we get $x_n\rightarrow x$ . What can you say about $f(x)$ .

**morito14** · September 11th 2010, 09:18 PM

Hum... nice explanation,

so $f(x)\subset K$ which is a contradiction, right?

so that would do it

**Jose27** · September 11th 2010, 09:27 PM

Originally Posted by morito14

Hum... nice explanation,

so $f(x)\subset K$ which is a contradiction, right?

so that would do it

Not really, since $f:K\rightarrow \mathbb{R}$ so $f(x) \in \mathbb{R}$ , the point is that $f(x) \leq \liminf_{x_n} f(x_n) = -\infty$ (and obviously $x\in K$ since K is compact (maybe this is what you meant?)).

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