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Let $f$ be a real-valued function(not necessarily continuous) on $\mathbb R^n$ with the property that for any $x\in \mathbb R^n $ and $\epsilon>0$, there is $\delta>0$ such that $f(y)>f(x)-\epsilon$ for all $y\in \mathbb R^n$ with $||y-x||<\delta$. Show that if $\lim_{||x||\to\infty}f(x)=1\ge f(c)$ for some $c \in \mathbb R^n$, then $f$ has a minimum value on $\mathbb R^n$.

I have no idea how to deal with this question. If $f$ were continuous, I would try the solution here, but I can't because there is no statement that $f$ is continuous. What should I do?

bellcircle
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1 Answers1

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So your first condition is widely known as lower semi-continuity. A variant of extreme value theorem applies to such functions, see this:

Any lower semicontinuous function $f: X \to \mathbb{R}$ on a compact set $K \subseteq X$ attains a min on $K$.

However you need a compact domain. To achieve that in your case you have to realize that the condition

$$\lim_{\lVert x\rVert->\infty}f(x)=1$$

implies that there's a well defined (and still lower semi continous) induced function

$$F:\mathbb{S}^n\to\mathbb{R}$$ $$F(x)=f(x)$$ $$F(\infty)=1$$

where $\mathbb{S}^n$ is an $n$-dimensional sphere constructed as the one-point compactification of $\mathbb{R}^n$, i.e. $$\mathbb{S}^n=\mathbb{R}^n\cup\{\infty\}$$ This space is compact and thus $F$ has to achieve its minimum somewhere, possibly in $\infty$.

Now since $$F(\infty)=1\geq f(c)=F(c)$$

for some $c$ then we see that if minimum is in $\infty$ then $F(\infty)=F(c)$ and thus $c$ is a minimum as well. All in all the minimum has to be in $\mathbb{R}^n$ and thus $f$ has minimum as well (since it shares values with $F$). $\Box$

freakish
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  • As an alternative to using the compactification, the limit condition tells us that for all $q<1$, there exists $R>0$ such that $f(x)>q$ for all $x$ with $x>R$. If there exists $x_0$ with $f(x_0)<1$, pick $q=\frac{f(x_0)+1}2$ (and if not, we already have $f(c)=\min$). Then we can use the compact closed $R$-ball. -- This is less elegant, but would already work with a weaker condition such as $\liminf_{|x|\to\infty}f(x)=1\ge f(c)$. – Hagen von Eitzen Jun 21 '17 at 14:05
  • @HagenvonEitzen Thanks. Actually the $\liminf$ condition works with compactification as well. The $F$ function is actually always well defined and $\liminf$ condition is enough (or rather equivalent) for it to be lower semicontinous. – freakish Jun 21 '17 at 14:10