# extended real valued function bounded below and attains its minimal value

• Sep 11th 2010, 01:56 PM
morito14
extended real valued function bounded below and attains its minimal value
Let f(x) be a real-valued function (an extended real valued function) on a metric space (M; d). Then
$\displaystyle \lim\sup_{x\rightarrow y}f(x)=\lim_{\epsilon\rightarrow0}\sup_{0<d(x,y)<\ epsilon}f(x)$
and
$\displaystyle \lim\inf_{x\rightarrow y}f(x)=\lim_{\epsilon\rightarrow0}\inf_{0<d(x,y)<\ epsilon}f(x)$

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

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??
• Sep 11th 2010, 09:08 PM
Jose27
Dealing with the first one (lower semicontinuity), assume f is not bounded below, then there exists a sequence $\displaystyle (x_n) \subset K$ such that $\displaystyle f(x_n) \rightarrow -\infty$, but $\displaystyle K$ is compact, so it's sequentially compact, so without loss of generality we get $\displaystyle x_n\rightarrow x$. What can you say about $\displaystyle f(x)$.
• Sep 11th 2010, 09:18 PM
morito14
Hum... nice explanation,

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

so that would do it
• Sep 11th 2010, 09:27 PM
Jose27
Quote:

Originally Posted by morito14
Hum... nice explanation,

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

so that would do it

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