Weak* separation theorem and Goldstine's theorem

Recall a version of separation theorem for a Banach space is the following.

Theorem 1. Let $X$ be a Banach space. Let $C$ be a nonempty open convex subset of $X$ and $x_0 \in X \backslash C$. Then there exists $\ell_0 \in X^*$ such that $\sup_{x \in C} \Re \ell_0(x) < \Re \ell_0(x_0)$.

There is a separation theorem in the weak* topology setting.

Theorem 2. Let $X$ be a Banach space. Let $C$ be a nonempty weak*-closed convex subset of $X^*$ and $\ell_0 \in X^* \backslash C$. Then there exists $x_0 \in X$ such that $\sup_{\ell \in C} \Re \ell(x_0) < \Re \ell_0(x_0)$.
Proof: Since $X^* \backslash C$ is weak*-open, we can find $x_1, \ldots, x_n \in X$ and $\epsilon > 0$ such that $U = \{\ell \in X^* : |\ell(x_i)| < \epsilon, i = 1, \ldots, n\}$ satisfies $\ell_0 + U \subset X^* \backslash C$. This implies that $\ell_0$ does not lie in the convex open set $C + U$. By Theorem 1 there exists $\psi \in X^{**}$ such that $\Re \psi(\ell_0) > \sup_{\ell \in C + U} \Re \psi(\ell) \geq \sup_{\ell \in C} \Re \psi(\ell)$. We want to show that $\cap_{i = 1}^n \ker(x_i) \subseteq \ker(\psi)$, since then $\psi$ is a linear combination of $x_1, \ldots, x_n$, thus $x_0 = \psi \in X$ and we are done. Let $\ell \in X^*$ with $\ell(x_i)$ for all $1 \leq i \leq n$. Note that $t\ell \in U$ for all $t \in \mathbb{R}$. Fix any $c \in C$. Then $M := \sup_{f \in U} \psi(f) < \psi(\ell_0) - \psi(c)$ is finite. Now for all $t \in \mathbb{R}$, $\psi(t\ell) \leq M$, i.e. $\psi(\ell) \leq M/t$ for all $t > 0$ and $\psi(\ell) \geq M/t$ for all $t < 0$. Therefore $\psi(\ell) = 0$.

The following is an application of the weak* separation theorem. (Notation: $B_Y = \{y \in Y: \|y\| \leq 1\}$.)

Theorem (Goldstine). The weak* closure of $B_X$ in $X^{**}$ is $B_{X^{**}}$.
Proof: First, $B_X \subseteq B_{X^{**}}$ and $B_{X^{**}}$ is weak*-compact by Alaoglu's Theorem. Hence the weak* closure $\tilde{B}$ of $B_X$ is a subset of $B_{X^{**}}$. Suppose the inclusion is proper and choose $w \in B_{X^{**}} \backslash \tilde{B}$. Apply Theorem 2 to find an $\ell \in X^*$ and $\epsilon > 0$ such that $\sup_{\psi \in \tilde{B}} \Re \psi(\ell_0) \leq \Re w(\ell_0) - \epsilon$. In particular, $\Re \ell_0(x) \leq \Re w(\ell_0) - \epsilon$ for all $x \in B_X$. This implies that $|\ell_0(x)| \leq \Re w(\ell_0) - \epsilon$ for all $x \in B_X$. Replace $w$ by some $e^{-i\theta}w$, we have $|\ell_0(x)| \leq |w(\ell_0)| - \epsilon$ for all $x \in B_X$. But then $\|\ell_0\| \leq |w(\ell_0)| - \epsilon \leq \|w\|\|\ell_0\| - \epsilon = \|\ell_0\| - \epsilon$. This is a contradiction.

Reference:
Fabian, Habala, Hajek,  Santalucia, Pelant, Zizler - Functional Analysis and Infinite Dimensional Geometry

Positive matrices I: Positive linear functionals

Let $\mathbb{M}_n$ denotes the algebra of n by n complex matrices. A matrix $A \in \mathbb{M}_n$ is said to be positive semi-definite or just positive if any one of the following equivalent conditions holds:

1. $x^*Ax \geq 0$ for any $x \in \mathbb{C}^n$;
2. $A = X^*X$ for some $X \in \mathbb{M}_n$;
3. $A$ is Hermitian (i.e. $A = A^*$) and all its eigenvalues are nonnegative (i.e. $\sigma(A) \subseteq [0, \infty)$).

We write $A \geq O$ if $A$ is a positive matrix.

A linear subspace $\mathcal{S}$ of $\mathbb{M}_n$ is an operator system if it contains $I$ and closed under the adjoint operation. 

Let $\mathcal{S} \subseteq \mathbb{M}_n$ be an operator system. A linear map $\Phi: \mathcal{S} \rightarrow \mathbb{M}_k$ is said to be positive if $\Phi(A) \geq O$ whenever $A \geq O$.

Note that we can write any $X \in \mathcal{S}$ as $X = A + iB$ where $A, B \in \mathcal{S}$ are Hermitian matrices given by $A = \frac{1}{2}(X + X^*)$ and $B = \frac{1}{2i}(X - X^*)$. Further, if $Y \in \mathbb{M}_n$ is Hermitian, then $Y = P - N$ for some positive matrices $P, N \in \mathcal{S}$. Here $P = \frac{1}{2}(\|Y\|I + Y)$ and $N = \frac{1}{2}(\|Y\|I - Y)$. The above decomposition gives the following result.

Proposition. If $\Phi: \mathcal{S} \rightarrow \mathbb{M}_k$ is a positive linear map, then $\Phi(X^*) = \Phi(X)^*$ for any $X \in \mathcal{S}$.

We now consider positive linear functionals on an operator system $\mathcal{S}$. It turns out in this case positivity is equivalent to that the functional attains its norm at the identity.

Theorem. Let $\varphi$ be a linear functional on $\mathcal{S}$. Then $\varphi$ is positive if and only if $\|\varphi\| = \varphi(I)$.

Proof: (=>) Let $X \in \mathcal{S}$. It is enough to show that $|\varphi(X)| \leq \|X\|\varphi(I)$. By replace $X$ by $e^{-i\theta}X$ for some suitable $\theta \in \mathbb{R}$, we may assume that $\varphi(X) \geq 0$. Write $X = A + iB$ where Hermitian $A, B \in \mathcal{S}$ are given by $A = \frac{1}{2}(X + X^*)$ and $B = \frac{1}{2i}(X - X^*)$. Since $\varphi(A), \varphi(B) \in \mathbb{R}$, we have $\varphi(X) = \varphi(A)$. Now write $A = P - N$ where $P, N \in \mathcal{S}$ are positive and given by $P = \frac{1}{2}(\|A\|I + A)$ and $N = \frac{1}{2}(\|A\|I - A)$. Note that $O \leq P \leq \|P\|I$, so $\varphi(P) \leq \|P\|\varphi(I)$. Now we have $\varphi(X) = \varphi(A) = \varphi(P) - \varphi(N) \leq \varphi(P) \leq \|P\|\varphi(I) \leq \|A\|\varphi(I) \leq \|X\|\varphi(I)$.
(<=) WLOG, assume $\|\varphi\| = \varphi(I) = 1$. Let $X \in \mathcal{S}$ be positive. Let $a = \min \sigma(X), b = \max \sigma(X)$ and $J = [a, b] \supseteq \sigma(X)$. We are going to show that $\varphi(X) \in J$, hence $\varphi(X) \geq 0$. Suppose not. Take a closed disk $D$ centered at $z_0$ with radius $r$ such that $J \subseteq D$ and $\varphi(X) \notin D$. It follows that $\sigma(X - z_0I) \subseteq  \{z: |z| \leq r\}$. Now $|\varphi(X) - z_0| = |\varphi(X - z_0I)| \leq \|\varphi\|\|X_0 - z_0I\| \leq r$. This contradicts the fact that $\varphi(X) \notin D$.

The above characterization immediately gives the following extension theorem for positive linear functionals.

Theorem (Krein Extension Theorem). Every positive linear functional on $\mathcal{S}$ extends to a positive  linear functional on $\mathbb{M}_n$.

Proof: Suppose $\varphi$ is a positive linear functional on $\mathcal{S}$. Then $\|\varphi\| = \varphi(I)$. By Hahn-Banach Theorem, $\varphi$ extends to a linear functional $\tilde{\varphi}$ on $\mathbb{M}_n$ with $\|\tilde{\varphi}\| = \|\varphi\| = \varphi(I) = \tilde{\varphi}(I)$, so $\tilde{\varphi}$ is positive.

Positive linear functionals on $\mathbb{M}_n$ have the following characterization.

Proposition. Let $\varphi$ be a positive linear functonal on $\mathbb{M}_n$. Then there exists a positive matrix $X \in \mathbb{M}_n$ such that $\varphi(A) = \mathrm{tr} AX$ for all $A \in \mathbb{M}_n$.

Proof: For $1 \leq i, j \leq n$, let $E_{ij} \in \mathbb{M}_n$ be the matrix with entry 1 at i-th row and j-th column and other entries zero and let $p_{ij} = \varphi(E_{ij})$. Take any $A = [a_{ij}] \in \mathbb{M}_n$. Then $\varphi(A) = \sum_{i, j} a_{ij}p_{ij} = \mathrm{tr}(AX)$ where $X = [x_{ij}]$ with $x_{ij} = p_{ji}$. Let $P = [p_{ij}]$, so that $X = P^*$. For any $x \in \mathbb{C}^n$, $x^*Xx = \sum_{i, j} \bar{x_i} x_{ij} x_j = \varphi(\sum_{i, j} \bar{x_i} E_{ji} x_j) = \varphi([\bar{x_j}x_i]) \geq 0$ since $\varphi$ and $[\bar{x_j}x_i] = xx^*$ are positive. Hence $X$ is positive.

Reference:
Bhatia - Positive Definite Matrices