3
$\begingroup$

Let $X$ be a topological space which is compact and connected.

$f$ is a continuous function such that;

$f : X \to \mathbb{C}-\{0\}$.

Explain why there exists two points $x_0$ and $x_1$ in $X$ such that $|f(x_0)| \le |f(x)| \le |f(x_1)|$ for all $x$ in $X$.

  • 0
    See also this question: [$X$ compact metric space, $f:X\rightarrow\mathbb{R}$ continuous attains max/min](http://math.stackexchange.com/questions/109548/x-compact-metric-space-fx-rightarrow-mathbbr-continuous-attains-max-min) or Corollary 3 in the ProofWiki article [Continuous Image of a Compact Space is Compact](http://www.proofwiki.org/wiki/Continuous_Image_of_a_Compact_Space_is_Compact).2012-07-05

3 Answers 3

2

the composite $X \to \mathbb{C} \setminus 0 \to \mathbb{R}_{> 0}$ given by first applying $f$ then the norm of a vector is a continuous map. Since $X$ is compact so is the image of this map as a subset of $\mathbb{R}_{>0}.$ Moreover by assumption on $X$ this set is connected. Connected compact subsets of $\mathbb{R}_{>0}$ are closed intervals. Then the claim follows.

  • 1
    that proof is very nice, thank you @mland, I made everything overcomplicated I guess because of the given properties.2012-07-04
2

Let $g(x)=|f(x)|$, observe that the complex norm is a continuous function from $\mathbb C$ into $\mathbb R$, therefore $g\colon X\to\mathbb R$ is continuous.

Since $X$ is compact and connected the image of $g$ is compact and connected. All connected subsets of $\mathbb R$ are intervals (open, closed, or half-open, half-closed); and all compact subsets of $\mathbb R$ are closed and bounded (Heine-Borel theorem).

Therefore the image of $g$ is an interval of the form $[a,b]$. Let $x_0,x_1\in X$ such that $g(x)=a$ and $g(x_1)=b$.

(Note that the connectedness of $X$ is not really needed, because compact subsets of $\mathbb R$ are closed and bounded, and thus have minimum and maximum.)

  • 0
    @Asaf: I'm just saying that a compact subset of a metric space is closed and bounded, and that is all that is used here, not the special fact about $\mathbb R^n$ that the converse holds, i.e. Heine--Borel or one half of it if you prefer. It is true that total boundedness also holds, and conversely if a metric space is complete and totally bounded then it is compact (at least assuming choice), but none of that seems very helpful here.2012-07-08
2

Define the function $g: X \to \mathbb{R} $ by $g(x) = |f(x)|$, which is continuous. Since X is compact, the result follows by the Extreme Value Theorem.