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Let $M$ be a topological space, $K\subset M$ compact, $f\colon M\to\mathbb{R}\cup\left\{+\infty\right\}$ lower semi-continious. Show that $f$ takes its minimum on $K$.

Good day,

we defined lower semi-continuity as follows: $f$ is lower semi-continious on $M$ if for any $x\in M$ and any $\alpha < f(x)$ there is a neighbourhood $U\subset M$ of $x$ such that $\alpha<f(y)$ for all $y\in U$.

So my idea is the following.

$f$ is lower semi-continious on $K$. Take any $x\in K$ and $\alpha < f(x)$, then there is a neighbourhood $U_x\subset M$ of $x$ such that $\alpha < f(y)~\forall y\in U$.

So it is $$ K\subset\bigcup_{x\in K}U_x. $$ Because of the compactness of $K$ there is a finite index set $I$ such that $$ K\subset\bigcup_{i\in I}U_{x_i}. $$ It is $f(y)>\alpha~\forall y\in U_{x_i}, i\in I$.

So for all $\alpha <\infty$ it is $f(x)>\alpha$ for all $x\in K$.

This means that it is $\inf_{x\in K}f(x)=-\infty$.

But do not know what to do with this result, is it helpful at all??

mathfemi
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  • I mean, no. $f : \mathbb R \to \mathbb R$ defined by $f(x) = |x|$ is continuous, hence lower-semi-continuous ; but $f$ does not take its minimum on the compact subset $[1,2]$ for instance. You need something more about $K$, or maybe you misformulated something. Do you mean that the infimum over $K$ of $f$ is actually a minimum? – Patrick Da Silva May 23 '14 at 15:13
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    $\alpha$ depends on $x.$ So you should write $\alpha_x.$ Since $K$ is compact you have $K\subset\bigcup_{i\in I}U_{x_i}$ with $I$ finite. Thus $f\geq \min_{i\in I}{\alpha_{x_i}},$ that is, $f$ is bounded from below. – mfl May 23 '14 at 15:16
  • @Patrick Da Silva The statement is, that because of the compactness of $K$ the lower semi-continious function $f$ takes its minimum on $K$. Another book says that then $f$ has a minimum of $K$. Do not know what is the better way to formulate it. – mathfemi May 23 '14 at 15:17
  • @mathfemi : The way I understand it my example is a counter-example to your statement. How is my example not a counter-example, if you believe you are right? Then perhaps I can understand your statement better. – Patrick Da Silva May 23 '14 at 15:18
  • Possible duplicate http://math.stackexchange.com/questions/123537/any-lower-semicontinuous-function-f-x-to-mathbbr-on-a-compact-set-k-sub – mfl May 23 '14 at 15:19
  • @ManuelFdzLpz Does Why is $f\geqslant\min_{i\in I}\left{\alpha_{x_i}\right}$? – mathfemi May 23 '14 at 15:21
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    Because, by definition of lower semi-continuity, in any $U_{x_i}$ you have $f(y)>\alpha_i\geq \min_{i\in I}{\alpha_i}$ – mfl May 23 '14 at 15:23
  • Ok, but the statement is not totally clear to me now: Set $b:=\min_{i\in I}\left{\alpha_i\right}$. Then for each $x\in K$ it is $f(x)>b$. Why does $f$ has its minimum on $K$ now? – mathfemi May 23 '14 at 15:35
  • @PatrickDaSilva The statement means that the restriction of $f$ to $K$ has a minimum. So your counterexample is not a counterexample; and this is good because the statement is true. – Etienne May 23 '14 at 15:41
  • I only see that $\inf_{x\in K}f(x)=b, b:=\min_{i\in I}\left{\alpha_{x_i}\right}$. Not that the inf is a minimum. – mathfemi May 23 '14 at 15:50

1 Answers1

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As mentioned by Manuel, the first step of your reasoning is wrong because the $\alpha$ depends on $x$.

However, your method of proof is quite correct, and with a minor adjustment it gives the statement you want. Here are the details.

Set $a:=\inf\{ f(x);\; x\in K\}$. This is well defined, but possibly equal to $-\infty$. It is enough to show that one can find $x$ such that $f(x)=a$. (In particular, this will show that $a>-\infty$).

Assume this is not true. Then, for any $x\in K$ you have $f(x)>\alpha$; so you can choose a real number $\alpha_x>a$ such that $f(x)>\alpha_x$.

Then your above reasoning (together with Manuel's comments) gives that one can find $x_1,\dots ,x_N\in K$ such that $$\forall x\in K\;: \; f(x)>\alpha:=\min (\alpha_{x_1},\dots ,\alpha_{x_N})\, . $$ It follows that $\inf\{ f(x);\; x\in K\}\geq\alpha$, i.e. $a\geq\alpha$; but this is a contradiction since $\alpha_{x_1},\dots ,\alpha_{x_N}$ are strictly greater than $a$ and hence $\alpha>a$.

Etienne
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