### The Krein-Milman Theorem

1. The Krein-Milman theorem in Locally Convex Spaces

My project work this semester focuses to understand the paper the Krein-Milman Theorem in Operator Convexity by Corran Webster and Soren Winkler, which appeared in the Transactions of the AMS [Vol 351, #1, Jan 99, 307-322]. But before reading the paper, it is imperative to understand the (usual) Krein-Milman theorem which is proved in the context of locally convex spaces. My understanding of this part follows the book A Course in Functional Analysis by J B Conway. To begin with we shall collect the preliminaries that we shall need to understand the Krein-Milman theorem.

1.1. Convexity

Let ${\mathbb{K}}$ denote the real(${\mathbb{R}}$) or the complex(${\mathbb{C}}$) number fields. Let ${X}$ be a vector space over ${\mathbb{K}}$. A subset of a vector space is called convex if for any two points in the subset, the line segment joining them lies completely in the subset. We make this idea more precise.

If ${a,b\in X}$, the line segment from ${a}$ to ${b}$ is given by the set

$\displaystyle [a,b]=\{(1-t)a+tb :0\leq t\leq 1\}.$

By an open line segmentwe shall mean that the end points should be excluded, that is

$\displaystyle (a,b)=\{(1-t)a+tb :0

We shall call a line segment proper if ${a\neq b}$. A subset ${V\subset X}$ is called convexif for any ${a,b\in V}$, the line segment ${[a,b]\subset V}$. It is easy to see that intersection of arbitrary number of convex subsets is again convex. We characterize a convex set as –

Proposition 1.1 : ${V}$ is convex if and only if whenever ${x_1,...,x_n\in V}$, and ${t_1,...,t_n\in [0,1]}$ with ${\sum_i t_i=1}$, then ${\sum_i t_ix_i\in V}$.

Proof: If the latter condition holds, then it is easy to see that ${V}$ is convex. For the converse part, we use induction on the number of points, ${n}$. For ${n=1}$, this is trivially true, while the case ${n=2}$ follows from the definition. Moreover note that if ${t_1=1}$, the theorem holds. So we shall assume ${t_1\neq 1}$. Suppose the theorem holds for ${k=n>2}$. Then, if there are ${n+1}$ points, say ${x_1,...,x_{n+1}}$, and ${t_1,...,t_{n+1}}$, with ${t_1\neq 1}$ and ${\sum_{i=1}^{n+1}t_i=1}$ then we can write

$\displaystyle t_1x_1+\sum_{i=2}^{n+1}t_ix_i=t_1x_1+(1-t_1)\sum_{i=2}^{n+1}\frac{t_i}{1-t_1}x_i.$

Now ${\sum_{i=2}^{n+1}\frac{t_i}{1-t_1}x_i\in V}$ by the induction hypothesis, and by the convexity of ${V}$, we get that ${\sum_{i=1}^{n+1}t_ix_i\in V}$. Thus the proof is complete. $\Box$

If ${V\subset X}$, the convex hull of ${V}$, denoted by ${\text{co}\;(V)}$, is the intersection of all convex sets containing ${V}$. Clearly this definition is meaningful since ${X}$ itself a convex set containing ${V}$, and that convexity is preserved under intersection. ${\text{co}\;(A)}$ is also convex. Using proposition 1.1, we have the following characterization of the convex hull of ${V}$.

The convex hull of ${V}$ is the set of all convex combinations of elements of ${V}$, that is

$\displaystyle co(A)=\left\lbrace t_1x_1+...+t_nx_n: x_i\in V,\; 0\leq t_i\leq 1,\; \sum_{i=1}^nt_i=1,\; n\; \text{is arbitrary}\right\rbrace$

As an example, the convex hull of three points ${x_1,x_2,x_3\in\mathbb{R}^2}$ is the closed triangular region formed by those points.

Definition 1. Let ${V}$ be a convex subset of a vector space ${X}$. A point ${a\in V}$ is called an extreme point of ${V}$, if there does not exist any proper open line segment containing ${a}$ and which lies completely in ${V}$. We denote the set of extreme points of ${V}$ by ${\text{ext}\;(V)}$.

Convex sets may or may not have extreme points. For example the set ${\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}}$, does not have any extreme point, whereas the extreme points of the set ${\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\}}$ are all the boundary points. The extreme points of a closed square region in ${\mathbb{R}^2}$ are its four corners.

We have following equivalent characterizations.

Proposition 1.3 : Let ${V}$ is convex subset of a vector space ${X}$, and let ${a\in V}$. Then the following are equivalent –

• (a) ${a\in\text{ext}\;V }$.
• (b) If ${x_1,x_2\in V}$ and ${0 and ${a=tx_1+(1-t)x_2}$, then either ${x_1\notin V}$ or ${x_2\notin V}$ or ${x_1=x_2=a}$.
• (c) If ${x_1,...,x_n\in V}$ and ${a\in\text{co}\;\{x_1,...,x_n\}}$, then ${a=x_k}$ for some ${k}$.
• (d) ${V-\{a\}}$ is a convex set.

Proof:The equivalence of ${(a)}$ and ${(b)}$ is straightforward from the definition of extreme point and convexity of ${V}$.

${(a)\Rightarrow (c)}$ : Let ${a\in\text{co}\;\{x_1,...,x_n\}}$. Then by proposition 1.2, we get that ${a=t_1x_1+...+t_nx_n}$, where ${0\leq t_i\leq 1}$, and ${\sum t_i=1}$. If ${t_i=1}$ for some ${i}$, we get that ${a=x_i}$. So suppose that ${0 for each ${i}$. Then we can write

$\displaystyle a=t_1x_1+(1-t_1)\sum_{i=2}^{n}\frac{t_i}{1-t_1}x_i$

This happens only if ${(1-t_1)x_1=t_2x_2+...+t_nx_n}$ which means ${a=x_1}$.

${(c)\Rightarrow (a)}$ : Suppose ${a\notin\text{ext}\;V}$. Then there are points ${x,y\in V}$ such that ${a=tx+(1-t)y}$, where ${0, and ${x\neq y}$. So ${a\in\text{co}\;\{x,y\}}$. Then ${(c)}$ dictates that ${a=x}$ or ${a=y}$. But then ${x=y}$.

${(a)\Rightarrow (d)}$ : We take two points ${x,y\in V-\{a\}}$, then ${\{tx+(1-t)y:0\leq t\leq 1\}\subset V-\{a\}}$, since ${a\neq tx+(1-t)y}$ for any ${t\in (0,1)}$. Thus ${V-\{a\}}$ is convex.

${(d)\Rightarrow (a)}$ : Suppose that ${a}$ is not an extreme point. Then there exists ${t\in (0,1)}$ and ${x,y\in V}$ with ${x\neq y}$, and that ${a=tx+(1-t)y}$. But this means that ${[x,y]}$ does not completely in ${V-\{a\}}$, which contradicts its convexity. $\Box$

We shall return to convexity once again after we have introduced the concept of a topological vector space.

1.2. Locally Convex Spaces

For doing analysis on vector spaces we need to have a topological structure on the space. Usually we deal with normed spaces, which already have a topology determined by the metric induced by the norm. We generalise to topological vector spaces which contain normed spaces as a subclass. We note, however, that the topological structure should be compatible with operations on the vector space. Thus we have the following definition.

Definition 2 : A topological vector space (TVS) is a vector space with a topology such that relative to this topology the maps defined by ${(x,y)\mapsto (x+y)}$ and ${(\alpha,x)\mapsto\alpha x}$ are continuous for all ${x,y\in X}$, and ${\alpha\in\mathbb{K}}$.

Note that ${X\times X}$ inherits the product topology. Addition is continuous means that if ${x_1,x_2\in X}$ and if ${V}$ is a neighborhood of ${x_1+x_2}$, there should exist neighborhoods ${V_1}$ of ${x_1}$ and ${V_2}$ of ${x_2}$ such that ${V_1+V_2\subset V}$. Similarly, if ${x\in X}$, and ${\alpha\in\mathbb{K}}$, and ${V}$ is a neighborhood of ${\alpha x}$ then for some ${r>0}$, and for some neighborhood ${W}$ of ${x}$, we have ${\beta W\subset V}$ whenever ${|\beta-\alpha|. We also note that translation by a vector and multiplication by a scalar are homeomorphisms on ${X}$.

Every normed space is a TVS, since the norm satisfies ${\|x+y\|\leq \|x\|+\|y\|}$ and ${\|\alpha x\|=|\alpha\|x\|}$. Now that we have a topology on the vector space, we can talk about open and closed sets. Let ${X}$ be a TVS. We define the closed convex hull of ${V}$, which is defined as the intersection of all closed convex subsets of ${X}$ containing ${V}$, and is denoted by ${\overline{\text{co}}\;(V)}$. Then ${\overline{\text{co}}\;(V)}$ is closed and convex. One may ask – whether ${\overline{\text{co}\;(V)}=\overline{\text{co}}\;(V)}$? This proposition answers that in affirmative.

Proposition 1.4 Let ${X}$ be a TVS and let ${V}$ be a convex subset of ${X}$. Then

• (a) ${\overline{V}}$ is convex, and
• (b) if ${a\in \text{int}\;V}$ and ${b\in\overline{V}}$, then

$\displaystyle [a,b)=\{tb+(1-t)a:0\leq t<1\}\subset\text{int}\;V.$

Proof:

• (a) Let ${a,b\in\overline{V}}$. Choose an element ${x\in V}$, and let ${t\in [0,1]}$. Let ${\{a_i\}}$ be a net in ${V}$ converging to ${a}$. Then since addition and scalar multiplication is continuous, we get

$\displaystyle (1-t)x+ta_i\rightarrow (1-t)x+ta,$

that is ${[x,a]\subset\overline{V}}$. Now let ${\{b_i\}}$ be a net in ${V}$ converging to ${b}$, then ${[b_i,a]\subset\overline{V}}$ for each ${i}$. Since ${\overline{V}}$ is closed, we have

$\displaystyle (1-t)b_i+ta\rightarrow (1-t)b+ta\in\overline{V}.$

Thus ${[b,a]\subset\overline{V}}$. Hence ${\overline{V}}$ is closed.

• (b) For this part, we fix ${t\in (0,1)}$, and ${c=tb+(1-t)a}$, where ${a\in \text{int}\;V}$ and ${b\in\overline{V}}$. Since translation is a homeomorphism, there is an open set ${W\subset X}$, such that ${a+W\subset V}$. Then for any ${d\in V}$

$\displaystyle V\supset td+(1-t)(a+W)=t(d-b)+tb+(1-t)(a+W)=[t(d-b)+(1-t)W]+c.$

It only remains to show that there is an element ${d\in V}$, such that ${0\in [t(d-b)+(1-t)W]}$ for then this set becomes a neighborhood of 0, and hence its translation to ${c}$ will mean that ${c\in\text{int}\; V}$. Now, ${0\in [t(d-b)+(1-t)W]}$ means ${d\in b-t^{-1}(1-t)W}$. But ${0\in -t^{-1}(1-t)W}$, and this set is open. Since ${b\in\overline{V}}$, we can always find ${d\in V}$.

$\Box$

Corollary – If ${V\subset X}$, then ${\overline{\text{co}\;(V)}=\overline{\text{co}}\;(V)}$.

Proof: Since ${\text{co}\;V}$ is convex, ${\overline{\text{co}\;V}}$ is convex too. Since ${\overline{\text{co}}\;V}$ is the intersection of all closed convex sets containing ${V}$, we have ${\overline{\text{co}}\;V\subset\overline{\text{co}\;V}}$. Now let ${x\in\overline{\text{co}\;V}}$, and let ${W}$ be any closed convex set containing ${V}$. Then clearly, ${W\supset\text{co}\;V}$. And since ${W}$ is closed, we also have ${W\supset\overline{\text{co}\;V}}$. Thus ${x\in W}$, and hence ${x\in\overline{\text{co}}\;V}$. $\Box$

Definition 3. Let ${X}$ be a vector space over ${\mathbb{K}}$. A seminorm is a function ${p:X\rightarrow [0,\infty)}$ satisfying –

• (a) ${p(x+y)\leq p(x)+p(y)}$ and
• (b) ${p(\alpha x)=|\alpha|p(x)}$

for all ${\alpha\in\mathbb{K}}$, and ${x,y\in X}$.

Note that, from (b), we have ${p(0x)=0p(x)=0}$, and thus ${p(0)=0}$. It may happen that ${p}$ is zero for elements which are not zero. A seminorm is called a norm if ${p(x)=0}$ if and only if ${x=0}$. Thus, every norm is a seminorm. Let ${X}$ be the set of all continuous functions ${f:\mathbb{R}\rightarrow\mathbb{R}}$. Fix ${x\in\mathbb{R}}$. Define ${p:X\rightarrow [0,\infty)}$, by ${p(f)=|f(x)|}$. Then

$\displaystyle p(f_1+f_2)=|(f_1+f_2)(x)|\leq |f_1(x)|+|f_2(x)|=p(f_1)+p(f_2).$

Also ${p(\alpha f)=|\alpha f(x)|=|\alpha|p(f)}$. Thus ${p}$ is a seminorm. But, ${p(f)=f(x)=0}$ does not mean that ${f=0}$.

Let ${X}$ be a vector space and ${\mathcal{P}}$ be a family of seminorms on ${X}$. Let ${T}$ be the topology on ${X}$ that has as subbase, the sets of the form

$\displaystyle \{x:p(x-x_0)<\epsilon\},$

where ${p\in\mathcal{P}, x_0\in X}$ and ${\epsilon>0}$. Since the basis determined by a subbase is the collection of all finite intersection of the elements of the subbase, we see that a subset ${U}$ is open if and only if for every ${x_0\in U}$ there are ${p_1,...,p_n\in\mathcal{P}}$ and ${\epsilon_1,...,\epsilon_n>0}$ such that

$\displaystyle \bigcap_{j=1}^n\left\lbrace x\in X:p_j(x-x_0)<\epsilon_j\right\rbrace\subset U.$

Then

Proposition 1.6 – ${X}$ with the above mentioned topology is a TVS.

Proof: We need to show that the maps ${(x,y)\mapsto (x+y)}$ and ${(\alpha,x)\mapsto \alpha x}$ are continuous for all ${x,y\in X}$, and ${\alpha\in\mathbb{K}}$.

To show the continuity of addition, let ${x,y\in Z}$, and let ${V}$ be a neighborhood of ${x+y}$. Then we need to find neighborhoods ${U_1}$ of ${x}$ and ${U_2}$ of ${y}$ such that ${U_1+U_2\subset V}$. Since ${V}$ is open there exist ${p_1,...,p_n\in\mathcal{P}}$ and ${\epsilon_1,...,\epsilon_n>0}$, such that the basis element

$\displaystyle B=\bigcap_{j=1}^n\left\lbrace x\in X:p_j(x-x_0)<\epsilon_j\right\rbrace\subset V.$

Let ${\epsilon=\min\{\epsilon_1,...,\epsilon_n\}}$, and define

$\displaystyle U_1=\bigcap_{j=1}^n\left\lbrace x\in X:p_j(z-x)<\frac{\epsilon}{2}\right\rbrace$

and

$\displaystyle U_2=\bigcap_{j=1}^n\left\lbrace x\in X:p_j(z-y)<\frac{\epsilon}{2}\right\rbrace$

as basis neighborhoods of ${x}$ and ${y}$ respectively. Let ${z_1\in U_1}$ and ${z_2\in U_2}$, then

$\displaystyle p_j(z_1+z_2-(x+y))\leq p_j(z_1-x)+p_j(z_2-y)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\leq\epsilon_j$

for each ${j=1,...,n}$. Thus ${z_1+z_2\in B\subset V}$. Thus ${U_1+U_2\subset V}$. The continuity of multiplication is handled in a similar manner. $\Box$

Definition 4 – A locally convex space (LCS) is a topological vector space whose topology is defined by a family of seminorms ${\mathcal{P}}$ such that

$\displaystyle \bigcap_{p\in\mathcal{P}}\left\lbrace x:p(x)=0 \right\rbrace=\{0\}.$

The condition imposed above ensures that the topology so defined is Hausdorff. To show that this is Hausdorff, let ${x,y\in X}$, with ${x\neq y}$. Then there is a ${p\in\mathcal{P}}$ for which ${p(x-y)\neq 0}$. For if ${p(x-y)=0}$ for all ${p\in\mathcal{P}}$, then ${x-y\in \bigcap_{p\in\mathcal{P}}\left\lbrace x:p(x)=0 \right\rbrace}$, which contradicts that only 0 belongs to that set. Let ${p(x-y)>\epsilon>0}$, and

$\displaystyle \begin{array}{rcl} U &=& \{z:p(x-z)<\frac{1}{2}\epsilon\} \\ V &=& \{z:p(y-z)<\frac{1}{2}\epsilon\} \end{array}$

Then ${U\cap V=\emptyset}$. For if ${z\in U\cap V}$, then

$\displaystyle p(x-y)\leq p(x-z)+p(y-z)<\epsilon,$

There is another characterization of a LCS, which explains why the name locally convex is given. A set ${V\subset X}$ is called balanced if ${\alpha x\in V}$, whenever ${x\in V}$, and ${|\alpha|\leq 1}$.

Theorem 1 Let ${X}$ be a TVS, and let ${\mathcal{U}}$ be the collection of all open convex balanced subsets of ${X}$. Then ${X}$ is locally convex if and only if ${\;\mathcal{U}}$ is a basis for the neighborhood system at 0.

1.3. The Krein-Milman Theorem

Roughly speaking, the Krein-Milman theorem is a generalization of the fact that in ${\mathbb{R}^n}$, one can recover the shape of a convex polygon once its extreme points are known.

We shall be using the following crucial separation theorem.

Theorem 2 Let ${X}$ be a complex LCS and let ${A}$ and ${B}$ be disjoint closed convex subsets of ${X}$. If ${B}$ is compact, then there is continuous linear functional ${f:X\rightarrow\mathbb{C}}$, an ${\alpha\in\mathbb{R}}$, and an ${\epsilon>0}$ such that for ${a\in A}$, and ${b\in B}$ we have

$\displaystyle \text{Re}\;f(a)\leq\alpha<\alpha+\epsilon\leq\text{Re}\; f(b).$

Theorem 3 If ${K}$ is a non-empty compact convex subset of a locally convex space ${X}$, then ${\text{ext}\;K\neq\emptyset}$ and ${K=\overline{\text{co}}(\text{ext}\;K)}$.

Proof: We shall prove the theorem in steps. First note that if ${K=\{x_0\}}$, a singleton set, then the theorem trivially holds. Therefore, we’ll assume that ${K}$ contains at least two points.

Step 1. According to Proposition 1.3, ${a\in K}$ is an extreme point if and only if ${K-\{a\}}$ is convex. Every locally convex space is Hausdorff, and since ${K}$ is a compact subset of ${X}$, we get that ${K}$ is closed. Hence ${K-\{a\}}$ is open relative to ${K}$. Thus, ${a\in K}$ is an extreme point if and only if ${K-\{a\}}$ is relatively open convex subset of ${K}$. In this first step of the proof, we shall establish the existence of a maximal relatively open convex proper subset of ${K}$ (using Zorn’s lemma).

Let ${\mathcal{U}}$ denote the collection of all relatively open convex proper subsets of ${K}$. This collection is not empty. For, let ${x_0\in K}$, then ${K-\{x_0\}}$ is open and non-empty (since we assumed that ${K}$ contains at least two points); let ${x_1\in K-\{x_0\}}$. Since ${X}$ is locally convex, there is an open convex subset ${V\subset K-\{x_0\}}$ containing ${x_1}$. Then ${V\cap K}$ is a relatively open convex proper subset of ${K}$. We introduce a partial order relation on ${\mathcal{U}}$ by inclusion, and let ${\mathcal{U}_0}$ be a totally ordered subset of ${\mathcal{U}}$. Put

$\displaystyle U_0=\bigcup\left\lbrace U:U\in\mathcal{U}_0\right\rbrace$

Clearly, ${U_0}$ is open and since ${\mathcal{U}_0}$ is totally ordered, ${U_0}$ is convex. We only have to show that ${U_0}$ is proper. If ${U_0=K}$, then ${\mathcal{U}_0}$ is an open cover for ${K}$, and by compactness of ${K}$, we have that ${U=K}$ for some ${U\in\mathcal{U}_0}$, which contradicts that ${U\in\mathcal{U}}$. Thus this totally ordered set has a maximal element, and hence by Zorn’s lemma, ${\mathcal{U}}$ has a maximal element ${U}$. This completes our first step.

Step 2. In this step we shall prove that ${K-U}$ consists of a single point, thereby proving that ext ${K}$ is not empty. If ${x\in K}$, and ${0\leq t\leq 1}$, let ${T_{x,t}:K\rightarrow K}$ be defined by

$\displaystyle T_{x,t}(y)=ty+(1-t)x.$

Clearly, ${T_{x,t}}$ is continuous and it preserves convex combinations, that is,

$\displaystyle T_{x,t}\left(\sum_{i=1}^n\lambda_iy_i\right)=\sum_{i=1}^n\lambda_iT_{x,t}(y_i),$

where ${y_i\in K}$ and ${\lambda_i\geq 0}$ for ${i=1,...,n}$ and ${\sum_{i=1}^ny_i=1}$. If we take ${x\in U}$, and ${0\leq t<1}$, then by convexity of ${U}$, we have ${T_{x,t}(U)\subset U}$, which implies ${U\subset T_{x,t}^{-1}(U)}$. Since pull back of an open set by a continuous function is open, we get that ${T_{x,t}^{-1}(U)}$ is open in ${K}$, and it is convex too. If ${y\in\overline{U}-U,}$ then by proposition 1.4 we have

$\displaystyle T_{x,t}(y)\in [x,y)\subset U.$

Thus ${\overline{U}\subset T^{-1}_{x,t}}$, and hence by maximality of ${U}$, we get ${T_{x,t}^{-1}(U)=K}$ which translates into ${T_{x,t}(K)\subset U}$.

Now, let ${V}$ be any open convex subset of ${K}$. Let ${y,z\in V\cup U}$. If both points lie in ${V}$, their convex combinations also lie in ${V}$, and hence in ${U\cup V}$. Suppose ${y\in U}$ and ${z\in V}$, then from what we just proved above, all the linear combinations ${(1-t)y+tz}$ (where ${0\leq t<1}$) lies in ${U}$. Thus ${[y,z]\subset U\cup V}$ which proves that ${U\cup V}$ is an open convex subset of ${K}$. But by maximality of ${U}$, we must have either ${U\cup V=U}$ or ${U\cup V=K}$. Using this we complete our claim that ${K-U}$ is singleton.

Let ${x,y\in K-U}$, and ${x\neq y}$. Let ${V_x}$ and ${V_y}$ be disjoint open convex subsets of ${K}$ such that ${x\in V_x}$ and ${y\in V_y}$. Since ${x\notin U}$, we have ${U\cup V_x=K}$ (from the previous paragraph). This implies that ${y\in V_y\subset K=U\cup V_x}$. But this is contradiction since ${y}$ does not lie in ${U}$ as well as ${V_x}$. Thus ${K-U}$ is singleton, and hence ext ${K}$ is not empty.

Step 3. In this last step we prove that ${K}$ is the closed convex hull of ext ${K}$, the last claim of the theorem. We first prove that if ${V}$ is an open convex subset of ${X}$, and ext ${K\subset V}$, then ${K\subset V}$. If this does not hold, there will be an open convex subset of ${V\subset X}$ such that ext ${K\subset V}$ but ${V\cap K\neq K}$. Then ${V\cap K\in\mathcal{U}}$, and is contained in a maximal element ${U}$. Since ${K-U=\{a\}}$ for some ${a\in\text{ext}\;K}$, this contradicts that ext ${K\subset V}$.

Finally, let ${E=\overline{\text{co}}(\text{ext}\;K)}$. Suppose ${E\neq K}$, and therefore let ${x\in K-E}$. Then ${E}$ and ${\{x\}}$ are disjoint closed convex sets with ${\{x\}}$ being compact. Then by theorem 1.8, there exists a continuous linear functional ${f:X\rightarrow\mathbb{C}}$, an ${\alpha\in\mathbb{R}}$, an ${\epsilon>0}$ such that ${E\subset\{z\in X:\text{Re}\;f(z)\leq\alpha\}}$, and ${x\in \{z\in X:\text{Re}\;f(z)\leq\alpha+\epsilon\}}$. Let ${V=\{z\in X:\text{Re}\;f(z)<\alpha+\frac{\epsilon}{2}\}}$. Then ${V}$ is open, convex and ${E\subset V}$ which implies ${\text{ext}\;K\subset V}$. But as we proved in the previous paragraph, we should have ${K\subset V}$. But this contradicts that ${\text{Re}\;f(x)\geq\alpha+\epsilon}$.

Hence ${K=E=\overline{\text{co}}(\text{ext}\;K)}$, and this completes the proof. $\Box$