Prove a set is closed using continuous function

Question

Suppose $U$ is an isometry isomorphism from $C(Q)$ to $C(K)$ where $C(Q)$ is the Banach space which contains all real value continuous function defined on $Q$ and its norm is the sup-norm, i.e. $\|f\|_{\infty}=\sup\{|f(a)|:a \in Q\}$

Define $Q$ and $K$ to be compact Hausdorff spaces.

Suppose $A =\{ t \in K:|(Uf)(t)|=\|Uf\|\}$. Prove that $A$ is closed in $K$.

This question is from here, whereby I still don't understand the explanation given. Can anyone explain why the set is closed in $K$?

Remark: Define a function $e:K\rightarrow \mathbb{R}$ given by $e(t)=|(Uf)(t)|$. Note that $\{\|Uf\| \}\subset \mathbb{R}$ is closed. If $e$ is continuous, then we have $A=e^{-1}(\{\|Uf\|\})$ is also closed. Now my question is, why $e$ is continuous?

What is $f$ in the definition of the set $A$? A fixed function in $C(Q)$? — gerw, Dec 23 '13 at 08:58

score 2 · Accepted Answer · answered Dec 23 '13 at 09:01

2

If I understand your setting correctly (see comment above), then $e$ is just a concatenation of the two continuous functions \begin{align*} |\cdot| &: \mathbb{R} \to \mathbb{R}, \\ U \, f &: K \to \mathbb{R}. \end{align*} Hence, $e$ is continuous.

answered Dec 23 '13 at 09:01

gerw

33,373

score 1 · Answer 2 · edited Dec 23 '13 at 09:02

1

By the (inverse) triangle inequality,

$$|e(t)-e(s)| = \big||(Uf)(t)|-|(Uf)(s)|\big| \leq |(Uf)(t)-(Uf)(s)|.$$

for ans $s,t \in K$. As $Uf \in C(K)$, this inequality implies the continuity of $e$.

edited Dec 23 '13 at 09:02

gerw

33,373

answered Dec 23 '13 at 08:59

saz

123,507

Prove a set is closed using continuous function

2 Answers2