Solution
= Solution
All <vector spaces> in this solution are complex, while <convex combinations> use real coefficients. An <extreme point> $x$ of a <convex set> $C$ is one for which $x=ty+(1-t)z$, $y,z\in C$ and $0<t<1$, forces $y=z=x$. Equivalently, $x$ is not the midpoint of two distinct points of $C$.
The <Krein-Milman theorem> says that every nonempty compact <convex set> $K$ in a Hausdorff <locally convex space> is the <closed convex hull> of its <extreme points>. Here is a proof. A <face of a convex set> $F\subseteq K$ is a <convex set> such that whenever an interior point of a segment in $K$ belongs to $F$, both endpoints belong to $F$. Consider nonempty compact faces, ordered by reverse inclusion. A chain has nonempty intersection by compactness and the <finite intersection property>; that intersection is again a compact face. The <Zorn lemma> therefore gives a minimal nonempty compact face $F$.
If $F$ contains distinct $x,y$, a continuous real <linear functional> $\ell$ on the underlying real <locally convex space> distinguishes them. Such a <linear functional> exists because the space is Hausdorff and locally convex, by the <Hahn-Banach theorem>. The maximizers of $\ell$ on $F$ form a nonempty proper compact face of $F$. A face of a face is a face of $K$, contradicting minimality. Therefore $F$ is a singleton, yielding an <extreme point>. The same argument inside any nonempty compact face of $K$ supplies an <extreme point> of $K$ lying in that face.
Let $D$ be the <closed convex hull> of the <extreme points> of $K$. It is a nonempty closed subset of compact $K$, hence compact. If $x_0\in K\setminus D$, the <Hahn-Banach separation theorem> provides a continuous real <linear functional> $\ell$ with $\ell(x_0)>\sup_D\ell$. Its maximizer set on $K$ is a nonempty compact face, which contains an <extreme point> $e$ of $K$. But $e\in D$ and $\ell(e)\geq\ell(x_0)>\sup_D\ell$, a contradiction. Thus \b[$\boxed{K=\overline{\operatorname{co}}(\operatorname{ext}K)}$], proving <Krein-Milman theorem>.
For a nonempty compact <Hausdorff space> $K$, the <extreme points of the dual unit ball of C(K)> are
$$
\boxed{\operatorname{ext}B_{C(K)^*}=\{\alpha\delta_x:x\in K,\ |\alpha|=1\},}
\qquad \delta_x(f)=f(x).
$$
This is the permitted description without proof, with $C(K)$ denoting the complex <space of continuous functions on a compact space> equipped with the <supremum norm>. If $K$ is empty, $C(K)=\{0\}$ and the <closed unit ball> of its <continuous dual space> has the single <extreme point> $0$ instead.
The complex <Banach–Stone theorem> states that a surjective complex-linear <isometric isomorphism of normed spaces> $T:C(K)\to C(L)$ has the form
$$
\boxed{(Tf)(y)=u(y)f(\varphi(y)),\qquad |u(y)|=1,}
$$
where $u\in C(L)$ and $\varphi:L\to K$ is a <homeomorphism>. Conversely, every such map is a surjective complex-linear <isometric isomorphism of normed spaces> for the <supremum norm>.
To prove this, suppose first that $K,L$ are nonempty. The <Banach-space adjoint> $T^*$ is a bijective <isometric isomorphism of normed spaces> on the <continuous dual spaces>, so it bijects their <closed unit balls> and preserves <extreme points>. The displayed <extreme point> description gives uniquely
$$
T^*\delta_y=u(y)\delta_{\varphi(y)},\qquad |u(y)|=1.
$$
Uniqueness follows by evaluating at the constant function $1$, and then using that <continuous functions> separate points of a compact <Hausdorff space>. Evaluation gives the required formula for $Tf$, and $u=T1$ is continuous. Surjectivity of $T^*$ on <extreme points> proves surjectivity of $\varphi$: the preimage of any $\delta_x$ is $\alpha\delta_y$ for some $y$. If $\varphi(y_1)=\varphi(y_2)$, every function in the range of $T$ has equal values at these two points after division by $u$; surjectivity of $T$ and separation of points force $y_1=y_2$. Thus $\varphi$ is bijective.
For every $f\in C(K)$, $f\circ\varphi=(Tf)/u$ is continuous. The evaluation map $K\to\mathbb C^{C(K)}$, $x\mapsto(f(x))_f$, is a continuous injection of a compact <Hausdorff space> into a Hausdorff <product topology>, hence a <homeomorphism> onto its image. Continuity of every coordinate $f\circ\varphi$ proves continuity of $\varphi$. A continuous bijection between compact <Hausdorff spaces> is a <homeomorphism>. Conversely the weighted-composition formula plainly preserves the <supremum norm>, and its inverse is
$$
(T^{-1}g)(x)=\frac{g(\varphi^{-1}(x))}{u(\varphi^{-1}(x))}.
$$
If one compact space is empty, a surjective <isometric isomorphism of normed spaces> forces the other to be empty, and the empty <homeomorphism> gives the corresponding trivial case. This completes <Banach–Stone theorem>.
Neither the <space of sequences converging to zero> $c_0$ nor $L^1[0,1]$ can be isometrically a <continuous dual space> of a <Banach space>. Indeed, every nonzero <continuous dual space> has a nonempty weak-star compact <closed unit ball> by <Banach-Alaoglu theorem>, and <Krein-Milman theorem> then guarantees an <extreme point>. A bijective linear <isometry> preserves <extreme points> of <closed unit balls>.
The <closed unit ball> of $c_0$ has no <extreme points>. Given $x\in c_0$ with $\|x\|_\infty\leq1$, choose $n$ with $|x_n|<1$ and $0<\varepsilon<1-|x_n|$. The two distinct elements $x\pm\varepsilon e_n$ remain in that <closed unit ball> and have midpoint $x$.
The <closed unit ball> of the complex <Lp space> $L^1[0,1]$ also has no <extreme points>. An element of <norm> less than one can be perturbed by a sufficiently small nonzero <Lp space> element in both directions. If $\|f\|_1=1$, the <non-atomic measure> $|f(t)|\,dt$ admits a measurable set $A$ of mass $1/2$; for example the continuous function $s\mapsto\int_0^s|f(t)|\,dt$ attains $1/2$. Put $h=f(\mathbf1_A-\mathbf1_{A^c})$. For $0<\varepsilon<1$, the distinct functions $f\pm\varepsilon h$ both have <Lp norm>
$$
(1+\varepsilon)\int_A|f|+(1-\varepsilon)\int_{A^c}|f|=1,
$$
with the two coefficients interchanged for the minus sign. Their midpoint is $f$. Thus \b[neither proposed space is isometrically a Banach dual].
Finally, $[0,1]$ is a <connected space>, whereas $[0,1]\cup[2,3]$ is disconnected. They cannot be homeomorphic. By <Banach–Stone theorem>, \b[$C[0,1]$ and $C([0,1]\cup[2,3])$ are not isometrically isomorphic] as complex <Banach spaces>.