Norm. Lin. Sp. is Isomet. Iso. to C(K) where K is a Comp. Haus. Sps

# Every Normed Linear Space is Isometrically Isomorphic to C(K) where K is a Compact Hausdorff Space

Theorem 1: Let $X$ be a normed linear space. Then there exists a compact Hausdorff space $K$ such that $X$ is isometrically isomorphic to $C(K)$. |

*Recall that a topological space $K$ is said to be Hausdorff if for every pair of distinct points $x, y \in K$ ($x \neq y$) there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$.*

*Here, "$C(K)$" denotes the space of continuous functions defined on $K$.*

**Proof:**Let $X$ be a normed linear space and let $B_*$ be the closed unit ball of the topological dual $X^*$ equipped with the weak* topology.

- By Alaoglu's Theorem, $B_*$ is compact. We aim to show that $B_*$ is Hausdorff. Let $\varphi, \psi \in B_*$ be such that $\varphi \neq \psi$. Then there exists a point $x_0 \in X$ such that $\varphi(x_0) \neq \psi(x_0)$. So choose $\epsilon > 0$ small enough so that:

\begin{align} \quad V_{\epsilon, \{ x_0 \}, \varphi} \cap V_{\epsilon, \{ x_0 \}, \psi} = \emptyset \end{align}

- So $B_*$ is Hausdorff.

- Now let $T : X \to C(K)$ be defined for all $x \in X$ by:

\begin{align} \quad T(x) = (J(x)) |_K \end{align}

- We have already proved that $T$ is linear and continuous. Furthermore, $T$ is isometric since for all $x \in X$ we have that:

\begin{align} \quad \| T(x) \| = \| (J(x)) |_K \| = \sup_{\varphi \in B_*} |J(x)(\varphi)| = \sup_{\varphi \in B^*} \| J(x) \| = \| x \| \end{align}

- So $X$ is isometrically isomorphic to $C(K)$. $\blacksquare$