Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been thinking about this problem for a long time right now, and feel stuck.

Given that $X$ is a topological space, and that for $f$ to be lower semicontinuous, for any $x \in X$ and $\epsilon > 0$, there is a neighborhood of $x$ such that $f(x) - f(x') < \epsilon$ for all $x'$ in the neighborhood of $x$. Also, for the compact definition, I can only think of using every open cover of $K$ has a finite subcover for the topological definition of compact sets.

I don't know where I went wrong, but I don't see where I used the fact of $K$ being compact. So I'll put what I did.

Suppose that $\inf f(K)$ exists, then let's call it $a \in \mathbb{R}$. By the intermediate value theorem, we know that there exists some $x_0 \in X$ such that $f(x_0) = a$. So given that $f$ is lower-semicontinuous, then for $x_0 \in X$ and $\epsilon > 0$, there is a neighborhood of $x_0$ such that $f(x_0) - f(x') < \epsilon$ for all $x'$ in the neighborhood of $x_0$. Note there is an open set in the neighborhood of $x_0$ that contains $x_0$. Then since $K$ is compact, there is a finite open subcover of $K$. So call this finite subcover of $K$, $J$ and call the open set that contains $x_0$, $L$. Hence, $J \cup L$ is an open cover of $K$ and $f$ attains a minimum on $K$.

If the infimum of $f(K)$ does not exist, then we can't consider the case above. Take any $c \in f(K)$. Consider the set of the form $f_c := \{x \in X | f(x) \leq c\}$. Then we see each $f_c$ is closed since $f$ being lower-semicontinuous implies $f^{-1}(-\infty,a]$ is closed for any $a \in \mathbb{R}$ and $f(K)$ being unbounded. For all $c \in f(K)$, we have that the intersection of all $f_c$ gives us that the intersection of the $f_c$ is closed. Not sure if this fact is in contrary of $K$ being compact. I got stuck here.

share|improve this question
    
Notice that lower semicontinuity is equivalent to $f(x)\leq \liminf_{y\to x} f(y)$ for all $x$. Take a minimizing sequence, use compactness and conclude. –  Jose27 Mar 23 '12 at 6:46
    
Hi Jose, the thing is i'm not allowed to use any other characterization of lower semicontinuity. But I'll take a look at it anyways for the heck of it. –  MathNewbie Mar 23 '12 at 7:03
    
@MathNewbie: Why aren't you allowed to use any characterization of lower semiconinuity? –  Beni Bogosel Mar 23 '12 at 7:04
    
I was informed by my instructor to strictly use the original definition I have, but I suppose that if I am to use the alternative characterization, it shouldn't be a problem given that I give a proof of their equivalence. At this point, I'll even try the alternative characterization of lower semicontinuity, since the definition that I've been given hasn't served me well or it's just me. –  MathNewbie Mar 23 '12 at 7:10
    
If this is homework you should add the homework tag. Do you have that compact subsets of $\mathbb{R}$ are closed and bounded, and that bounded sequences have convergent subsequences? –  Patrick Mar 23 '12 at 7:18
show 1 more comment

2 Answers

up vote 4 down vote accepted

Lower semicontinuity need not imply intermediate value property(IVP), and a function on a compact interval in $\mathbb{R}$ which satisfies IVP can fail to have the minimum.

But your $\inf = -\infty$ case proof can be elaborated to conclude that $f$ cannot be unbounded below. What you need is the finite intersection property of a compact set. Or, you can just consider the open cover (why?) formed by the open sets $U_c = \{ x : f(x) > c \}$ to obtain the boundedness of $f$.

Here is another possible approach:

Suppose $f$ has no minimum, and let $\alpha = \inf f(K)$. Then for each $x \in K$, we have $\alpha < f(x)$ and there is $\epsilon(x) > 0$ satisfying $$\alpha < f(x) - \epsilon(x).$$ Then by lower-semicontinuity, there exists a neighborhood $U(x)$ of $x$ such that $f(x) - \epsilon(x) < f(y)$ for all $y \in U(x)$. Now $U(x)$ covers $K$, so there are finitely many $x_1, \cdots, x_n \in K$ where $$K \subset U(x_1) \cup \cdots \cup U(x_n).$$ Let $\beta$ be given by $$ \beta = \min \{ f(x_k) - \epsilon(x_k) : k = 1, \cdots, n \}.$$ Then clearly $\alpha < \beta$, and for any $y \in K$, $y \in U(x_k)$ for some $k$ and hence $\beta \leq f(x_k) - \epsilon(k) < f(y)$, which means that $\beta \leq \inf f(K) = \alpha$, a contradiction! Therefore $f$ must attain its minimum at some point in $K$.

share|improve this answer
add comment

So this won't help the person who needed it for homework, but for future visitors, here's an approach which is similar to the one above, but uses a contradiction of compactness.

Suppose $f$ has no minimum in $K$, then $\neg(\exists x\in K)(\forall y\in K)(f(x)\leq f(y))$ which is equivalent to $(\forall x\in K)(\exists y\in K)(f(y)<f(x))$ which is equivalent to $(\forall x\in K)(\exists y\in K)(x\in f^{-1}((f(y), +\infty)))$.

But this means the collection of open sets $\{f^{-1}((f(x), +\infty)):x\in K\}$ is an open cover of $K$.

We claim this open cover has no finite subcover.

Suppose $\{x_1, \cdots, x_n\}\subset K$ is finite, and define $N = \displaystyle\arg\min_{1\leq k\leq n}f(x_k)$, then $\displaystyle\bigcup_{1\leq k\leq n}(f(x_k),+\infty) = (f(x_N), +\infty)$. Then we have $\displaystyle x_N\in K$, but $x_N \notin f^{-1}((f(x_N), +\infty))= f^{-1}(\bigcup_{1\leq k\leq n}(f(x_k),+\infty)) = \bigcup_{1\leq k\leq n}f^{-1}((f(x_k),+\infty))$ and so $\{f^{-1}((f(x_k),+\infty)):1\leq k \leq n\}$ doesn't cover $K$.

This contradicts the compactness of $K$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.