diff --git a/src/harmonic/rep/pos.tex b/src/harmonic/rep/pos.tex
new file mode 100644
index 0000000..cb1096b
--- /dev/null
+++ b/src/harmonic/rep/pos.tex
@@ -0,0 +1,233 @@
+\section{Positive-Definite Functions}
+\label{section:positive-definite-function}
+
+\begin{definition}[Positive Linear Functional]
+\label{definition:group-l1-positive-functional}
+    Let $G$ be a locally compact group and $\phi \in L^1(G; \complex)^*$\footnote{Due to technical difficulties, positive linear functionals are a priori not represented by $L^\infty$ functions. }, then $\phi$ is \textbf{positive} if $\dpn{f^* * f,  \phi}{L^1(G; \complex)} \ge 0$ for all $f \in L^1(G)$. 
+\end{definition}
+
+
+\begin{definition}[GNS Triple (Locally Compact Group)]
+\label{definition:gns-triple-lcg}
+    Let $G$ be a locally compact group and $\phi \in L^1(G; \complex)^*$. For each $f, g \in L^1(G; \complex)$, let
+    \[
+    \dpn{f, g}{\phi} = \dpn{g^* * f, \phi}{L^1(G; \complex)}
+    \]
+
+    Let $N_\phi = \bracsn{f \in L^1(G; \complex)| \dpn{f, f}{\phi} = 0}$, $H_\phi^0 = L^1(G; \complex)/N_\phi$, $H_\phi$ be its completion with respect to $\dpn{\cdot, \cdot}{\phi}$, and
+    \[
+    \pi_\phi^0: G \to B(H_\phi^0) \quad \pi_\phi^0(x)(f + N_\phi) = L_xf + N_\phi
+    \]
+
+    For each $x \in G$, let $\pi_\phi(x)$ be the continuous extension of $\pi_\phi^0(x)$ to an element of $B(H_\phi)$, then:
+    \begin{enumerate}
+        \item $H_\phi$ equipped with the continuous extension of $\dpn{\cdot, \cdot}{\phi}$ is a Hilbert space.
+        \item $(H_\phi, \pi_\phi)$ is a well-defined unitary representation of $G$. 
+    \end{enumerate}
+
+    Let $\pi_\phi: L^1(G; \complex) \to B(H_\phi)$ also denote the *-representation of $L^1(G; \complex)$ determined by $\pi_\phi$, then:
+    \begin{enumerate}[start=2]
+        \item For each $f, g \in L^1(G; \complex)$, $\pi_\phi(f)(g + N_\phi) = f * g + N_\phi$. 
+        \item There exists a cyclic vector $\xi_\phi \in H_\phi$ for $\pi_\phi$ such that for each $f, g \in L^1(G; \complex)$, 
+        \[
+        \dpn{f, g}{\phi} = \dpn{\pi_\phi(f)\xi_\phi, \pi_\phi(g)\xi_\phi}{H_\phi}
+        \]
+        \item For each $x \in G$, let $\Phi(x) = \dpn{\pi_\phi(x)\xi_\phi, \xi_\phi}{H_\phi}$, then $\Phi \in BC(G; \complex)$ and
+        \[
+        \dpn{f, \phi}{L^1(G; \complex)} = \int_G f(x)\Phi(x)dx
+        \]
+
+        for all $f \in L^1(G; \complex)$. 
+    \end{enumerate}
+
+    Moreover, 
+    \begin{enumerate}
+        \item[(U)] Let $(K, \tau)$ be a unitary representation of $G$ with cyclic vector $\eta$ such that $\dpn{\pi_\phi(x)\xi_\phi, \xi_\phi}{H_\phi} = \dpn{\tau(x)\eta, \eta}{K}$, then there exists a unitary equivalence $U: H_\phi \to K$ such that $U\xi_\phi = \eta$. 
+    \end{enumerate}
+    
+
+    The representation $(H_\phi, \pi_\phi)$ is the \textbf{unitary cyclic representation of $G$ induced by $\phi$}, and the triple $(H_\phi, \pi_\phi, \xi_\phi)$ is the \textbf{Gelfand-Naimark-Segal (GNS) triple associated with $\phi$}. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Theorem 3.20]{FollandHarmonic}}}. ]
+    (1): Since $\phi$ is positive-definite, $\dpn{\cdot, \cdot}{\phi}$ is a pseudo inner product, and $H_\phi$ equipped with its continuous extension is a Hilbert space. 
+    
+    (2): By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, for any $f, g \in L^1(G; \complex)$, 
+    \[
+    \dpn{f, g}{\phi} \le \dpn{f, f}{\phi}^{1/2} \cdot \dpn{g, g}{\phi}^{1/2}
+    \]
+
+    In particular, if $g \in N_\phi$, then
+    \begin{align*}
+    \dpn{f * g, f * g}{\phi} &= \dpn{g^* * f^* * f * g, \phi}{L^1(G; \complex)} \\
+    &= \dpn{g^** (f^**f*g), \phi}{L^1(G; \complex)} = \dpn{f^**f*g, g}{\phi}
+    \end{align*}
+    
+
+    so $N_\phi$ is a left ideal of $L^1(G; \complex)$. Let $x \in G$, then $(L_xg)^* = L_{-x}g^*$, and
+    \begin{align*}
+        \dpn{L_xf, L_xg}{\phi}  &= \dpn{(L_xg)^* * L_xf, \phi}{L^1(G; \complex)} = \dpn{(L_{-x}g^*) * (L_xf), \phi}{L^1(G; \complex)} \\
+        &= \dpn{g^* * f, \phi}{L^1(G; \complex)} = \dpn{f, g}{\phi}
+    \end{align*}
+
+    by \autoref{proposition:convolution-function-gymnastics}. Therefore $\pi_\phi^0(x)$ extends to a unitary map on $H_\phi$, and $\pi_\phi$ is a well-defined unitary representation of $G$, which is strong-operator continuous by \hyperref[continuity of translation]{proposition:haar-translation}.
+
+    (3): Let $f, g, h \in L^1(G; \complex)$, then after passing through \hyperref[Riemann sums]{lemma:convolution-distribute-functional}, 
+    \begin{align*}
+        \dpn{\pi_\phi(f)g + N_\phi, h + N_\phi}{H_\phi} &= \int_G f(x)\dpn{\pi_\phi(x)g + N_\phi, h + N_\phi}{H_\phi} dx \\
+        &= \int_G f(x)\dpn{L_xg, h}{\phi} dx = \dpn{f * g, h}{\phi} \\
+        &= \dpn{f * g + N_\phi, h + N_\phi}{H_\phi}
+    \end{align*}
+    
+    As the above holds for all $h \in L^1(G; \complex)$, $\pi_\phi(f)(g + N_\phi) = f * g + N_\phi$. 
+
+    (4): Using \autoref{proposition:group-algebra-approximate-identity}, let $\angles{\psi_\alpha}_{\alpha \in A} \subset L^1(G; \complex)$ be an approximate identity, and $\xi_\phi \in H_\phi$ be defined by
+    \[
+    \dpn{\xi_\phi, g + N_\phi}{H_\phi} = \lim_{\alpha \in A}\dpn{\psi_\alpha, g}{\phi} \quad \forall g \in L^1(G; \complex)
+    \]
+
+    then $\normn{\xi_\phi}_{H_\phi} \le \norm{\phi}_{L^1(G; \complex)^*}^{1/2}$. Moreover, for any $f, g \in L^1(G; \complex)$, 
+    \begin{align*}
+        \dpn{\pi_\phi(f)\xi_\phi, \pi_\phi(g)\xi_\phi}{H_\phi} &= \lim_{\alpha \in A}\dpn{\pi_\phi(f)(\psi_\alpha + N_\phi), \pi_\phi(g)\xi_\phi}{H_\phi} \\
+        &= \lim_{\alpha \in A}\dpn{f * \psi_\alpha + N_\phi, \pi_\phi(g)\xi_\phi}{H_\phi} \\
+        &= \dpn{f + N_\phi, \pi_\phi(g)\xi_\phi}{H_\phi} = \dpn{f + N_\phi, g + N_\phi}{H_\phi}
+    \end{align*}
+\end{proof}
+
+\begin{corollary}
+\label{corollary:group-l1-positive-function}
+    Let $G$ be a locally compact group and $\phi \in L^1(G; \complex)^*$ be a positive linear functional, then:
+    \begin{enumerate}
+        \item There exists $\Phi \in BC(G; \complex)$ such that
+        \[
+        \dpn{f, \phi}{L^1(G; \complex)} = \int_G f(x)\Phi(x) dx
+        \]
+
+        for all $f \in L^1(G; \complex)$. 
+        \item $\Phi(1_G) = \norm{\Phi}_u = \norm{\phi}_{L^1(G; \complex)^*}$.
+    \end{enumerate}
+\end{corollary}
+\begin{proof}
+    (1): Let $(H_\phi, \pi_\phi, \xi_\phi)$ be the \hyperref[GNS triple]{definition:gns-triple-lcg} associated with $\phi$. For each $x \in G$, let $\Phi(x) = \dpn{\pi_\phi(x)\xi_\phi, \xi_\phi}{H}$, then for any $f \in L^1(G; \complex)$
+    \[
+    \dpn{\pi_\phi(f)\xi_\phi, \xi_\phi}{H_\phi} = \int_G f(x)\dpn{\pi_\phi(x)\xi_\phi, \xi_\phi}{H_\phi}dx = \int_G f(x)\Phi(x)dx
+    \]
+
+    by definition of the representation. Since $\pi_\phi$ is strong-operator continuous, $\Phi \in C(G; \complex)$. 
+    
+    (2): By \autoref{theorem:lp-dual-function}, $\Phi \in BC(G; \complex)$ with $\norm{\Phi}_u = \norm{\phi}_{L^1(G; \complex)^*}$. In addition, for any $x \in G$, 
+    \[
+    |\Phi(x)| = |\dpn{\pi_\phi(x)\xi_\phi, \xi_\phi}{H_\phi}| \le \normn{\xi_\phi}_{H_\phi}^2 = \Phi(1_G)
+    \]
+\end{proof}
+
+\begin{lemma}
+\label{lemma:group-positive-computation-gymnastics}
+    Let $G$ be a locally compact group, $\phi \in L^\infty(G; \complex)$, and $f \in L^1(G; \complex)$, then
+    \[
+    \int_G (f^* * f) \phi = \iint_{G \times G} f(x)\ol{f(y)}\phi(y^{-1}x)dydx
+    \]
+\end{lemma}
+\begin{proof}
+    \begin{align*}
+        \int_G (f^* * f)(x)\phi(x) dx &= \iint_G \Delta_G(y^{-1})\ol{f(y^{-1})}f(y^{-1}x)\phi(x)dydx \\
+        &= \iint_G \ol{f(y)} f(yx)\phi(x)dy dx \\
+        &= \iint_{G \times G} f(x) \ol{f(y)} \phi(y^{-1}x)dydx
+    \end{align*}
+    
+\end{proof}
+
+
+
+\begin{definition}[Positive-Definite Function]
+\label{definition:group-positive-definite-function}
+    Let $G$ be a locally compact group and $\phi \in BC(G; \complex)$, then the following are equivalent:
+    \begin{enumerate}
+        \item The mapping $f \mapsto \dpn{f, \phi}{L^1(G; \complex)}$ is a positive linear functional on $L^1(G; \complex)$. 
+        \item For each $n \in \natp$, $\seqf{\lambda_j} \subset \complex$ and $\seqf{x_j} \subset G$, 
+        \[
+        \sum_{i = 1}^n \sum_{j = 1}^n \lambda_i \ol{\lambda_j} \phi(x_j^{-1}x_i) \ge 0
+        \]
+    \end{enumerate}
+    
+    If the above holds, then $\phi$ is \textbf{positive-definite}. The set $P(G)$ is the \textbf{space of positive-definite functions} on $G$, $P_0(G) = \bracsn{\phi \in P(G)|\norm{\phi}_u \le 1}$, and $S(G) = \bracsn{\phi \in P(G)|\norm{\phi}_u = 1}$.
+\end{definition}
+\begin{proof}[Proof, {{\cite[Proposition 3.35]{FollandHarmonic}}}. ]
+    (1) $\Rightarrow$ (2): Using \autoref{proposition:group-algebra-approximate-identity}, let $\angles{\psi_\alpha}_{\alpha \in A} \subset L^1(G; \complex)$ be an approximate identity such that for all $U \in \cn_G(1)$, there exists $\alpha_0 \in A$ with $\supp{\psi_\alpha} \subset U$ for all $\alpha \ge \alpha_0$. 
+    
+    For each $\alpha \in A$, $\seqf{\lambda_j} \subset \complex$, and $\seqf{x_j} \subset G$, let $f_\alpha = \sum_{j = 1}^n \lambda_j L_{x_j}\psi_\alpha$, then by continuity of $\phi$, 
+    \begin{align*}
+        0 \le \int_G (f^*_\alpha * f_\alpha) \phi &= \sum_{i, j = 1}^n \lambda_i \ol{\lambda_j} \iint_{G \times G} \psi_\alpha(x_i^{-1}y)\ol{\psi_\alpha(x_j^{-1}y)}\phi(z^{-1}y)dydz \\
+        0 \le \lim_{\alpha \in A}\int_G (f^*_\alpha * f_\alpha) \phi &= \sum_{i, j = 1}^n \lambda_i \ol{\lambda_j}\phi(x_j^{-1}x_i)
+    \end{align*}
+    
+
+    (2) $\Rightarrow$ (1): Let $f \in C_c(G; \complex)$, then $F(x, y) = f(x)\ol{f(y)}\phi(y^{-1}x)$ is uniformly continuous by \autoref{proposition:lcg-cc-uc}. As such, there exists a partition $K = \bigsqcup_{j = 1}^n U_j$ and $\seqf{x_j} \subset G$ such that $|F(x_i, x_j) - F(x, y)| \le \eps$ for all $(x, y) \in U_i \times U_j$ and $1 \le i, j \le n$. Thus
+    \[
+    \abs{\int_G (f^* * f) \phi - \sum_{i, j = 1}^n F(x_i, x_j) \cdot |U_i| \cdot |U_j|} \le \eps |K|^2
+    \]
+
+    As
+    \[
+    \sum_{i, j = 1}^n F(x_i, x_j) \cdot |U_i| \cdot |U_j| = \sum_{i, j = 1}^n f(x_i)|U_i| \cdot \ol{f(x_j)|U_j|} \cdot \phi(x_j^{-1}x_i) \ge 0
+    \]
+
+    and the above holds for all $\eps > 0$, $\int_G (f^* * f) \phi \ge 0$. By \hyperref[density of $C_c(G; \complex)$ in $L^1(G; \complex)$]{proposition:radon-cc-dense} and continuity of the convolution product, $\int_G (g^* * g) \phi \ge 0$ for all $g \in L^1(G; \complex)$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:gns-irreducible-lcg}
+    Let $G$ be a locally compact group, $\phi \in S(G)$ be a state, and $(H_\phi, \pi_\phi, \xi_\phi)$ be the GNS triple associated with $\phi$, then $(H_\phi, \pi_\phi)$ is irreducible if and only if $\phi$ is an extreme point of $S(G)$. 
+\end{proposition}
+\begin{proof}
+    ($\Rightarrow$): Suppose that $(H_\phi, \pi_\phi)$ is irreducible. Let $\psi_1, \psi_2 \in S(G)$ and $t \in (0, 1)$ such that $\phi = t\psi_1 + (1 - t)\psi_2$. Define
+    \[
+    T: H_\phi \to H_\phi \quad \dpn{T\pi_\phi(f)\xi_\phi, \pi_\phi(g)\xi_\phi}{H_\phi} = \dpn{f, g}{\psi_1} = \dpn{g^* * f, \psi_1}{L^1(G; \complex)}
+    \]
+
+    where $f, g \in L^1(G; \complex)$. Since both $\psi_1$ and $\psi_2$ are positive-definite functions, $\phi - t\psi_1$ is also positive-definite. In particular, $\dpn{f, f}{\psi_1} \le t^{-1}\dpn{f, f}{\phi}$ for all $f \in L^1(G; \complex)$, so $T$ extends continuously into a positive operator on $H_\phi$. 
+
+    For any $x \in G$ and $f, g \in L^1(G; \complex)$, 
+    \begin{align*}
+        \dpn{\pi_\phi(x)T\pi_\phi(f)\xi_\phi, \pi_\phi(g)\xi_\phi}{H_\phi} &= \dpn{T\pi_\phi(f)\xi_\phi, \pi_\phi(x^{-1})\pi_\phi(g)\xi_\phi}{H_\phi} \\
+        &= \dpn{T\pi_\phi(f)\xi_\phi, \pi_\phi(L_{x^{-1}}g)\xi_\phi}{H_\phi} \\
+        &= \dpn{f, L_{x^{-1}}g}{\psi_1} = \dpn{(L_{x^{-1}}g)^* * f, \psi_1}{L^1(G; \complex)} \\
+        &= \dpn{g^* * (L_xf), \psi_1}{L^1(G; \complex)} = \dpn{L_xf, g}{\psi_1} \\
+        &= \dpn{T\pi_\phi(L_xf)\xi_\phi, \pi_\phi(g)\xi_\phi}{H_\phi} \\
+        &= \dpn{T\pi_\phi(x)\pi_\phi(f)\xi_\phi, \pi_\phi(g)\xi_\phi}{H_\phi}
+    \end{align*}
+
+    so $T \in \pi_\phi(G)'$. By \hyperref[Schur's Lemma]{lemma:lcg-schur}, there exists $\lambda \in \complex \setminus \bracsn{0}$ such that $T = \lambda I$. 
+
+    Using \autoref{proposition:group-algebra-approximate-identity}, let $\angles{g_\alpha}_{\alpha \in A} \subset L^1(G; \complex)$ be an approximate identity for $L^1(G; \complex)$, then for any $f \in L^1(G; \complex)$, 
+    \[
+    \dpn{f, \psi_1}{L^1(G; \complex)} = \lim_{\alpha \in A}\dpn{f, g_\alpha}{\psi_1} = \lim_{\alpha \in A}\lambda\dpn{f, g_\alpha}{\phi} = \lambda \dpn{f, \phi}{L^1(G; \complex)}
+    \]
+
+    Since $\phi, \psi_1 \in S(G)$, $\lambda = 1$, and $\phi = \psi_1$. As the argument is symmetric, $\phi = \psi_1 = \psi_2$. 
+
+    ($\Leftarrow$): Suppose that $(H_\phi, \pi_\phi)$ is reducible. By \hyperref[Schur's Lemma]{lemma:lcg-schur}, $\pi_\phi(G)'$ is non-trivial. Since $\pi_\phi(G)'$ is a von Neumann algebra, it admits a non-trivial projection $P \in B(H_\phi)$ by \autoref{theorem:vn-projection-norm-dense}. 
+    
+    Let $\xi_1 = P\xi_\phi/\normn{P\xi_\phi}_{H_\phi}$ and $\xi_2 = (1 - P)\xi_\phi/\normn{\xi_\phi - P\xi_\phi}_{H_\phi}$. For each $f \in L^1(G; \complex)$, let
+    \[
+    \dpn{f, \psi_1}{L^1(G; \complex)} = \dpn{\pi_\phi(f)\xi_1, \xi_1}{H_\phi} \quad
+    \dpn{f, \psi_2}{L^1(G; \complex)} = \dpn{\pi_\phi(f)\xi_2, \xi_2}{H_\phi}
+    \]
+
+    then $\psi_1, \psi_2 \in S(G)$. Since $P \in \pi_\phi(G)'$, for any $f \in L^1(G; \complex)$, 
+    \begin{align*}
+        \dpn{f, \phi}{L^1(G; \complex)} &= \dpn{\pi_\phi(f)\xi_\phi, \xi_\phi}{H_\phi} \\
+        &= \dpn{\pi_\phi(f)P\xi_\phi, P\xi_\phi}{H_\phi} + \dpn{\pi_\phi(f)(I - P)\xi_\phi, (I - P)\xi_\phi}{H_\phi} \\
+        &= \normn{P\xi_\phi}_{H_\phi}^2\dpn{\pi_\phi(f)\xi_1, \xi_1}{H} + \normn{(1 - P)\xi_\phi}_{H_\phi}^2 \dpn{\pi_\phi(f)\xi_2, \xi_2}{H} \\
+        &= \normn{P\xi_\phi}_{H_\phi}^2 \dpn{f, \psi_1}{A} + \normn{(1 - P)\xi_\phi}_{H_\phi}^2 \dpn{f, \psi_2}{L^1(G; \complex)}
+    \end{align*}
+  
+    so $\phi$ is a strict convex combination of two states. If $\psi_1 = \phi$, then for any $f, g \in L^1(G; \complex)$, 
+    \begin{align*}
+        \dpn{\pi_\phi(f)\xi_\phi, \pi_\phi(g)\xi_\phi}{H_\phi} &= \normn{P\xi_\phi}_{H_\phi}^{-2} \dpn{P\pi_\phi(f)\xi_\phi, \pi_\phi(g)\xi_\phi}{H_\phi} \\
+        \dpn{[\pi_\phi(f) - \normn{P\xi_\phi}_{H_\phi}^{-2}P\pi_\phi(f)]\xi_\phi,\pi_\phi(g) \xi_\phi}{H_\phi} &= 0
+    \end{align*}
+
+    so $I - \normn{P\xi_\phi}_{H_\phi}^{-2}P = 0$, which contradicts the fact that $P$ is a non-trivial projection.
+\end{proof}
+
+
diff --git a/src/op/c-star/gns.tex b/src/op/c-star/gns.tex
index 0014c1c..e850969 100644
--- a/src/op/c-star/gns.tex
+++ b/src/op/c-star/gns.tex
@@ -61,9 +61,9 @@
 
     For each $x \in A$, let $\pi_\phi(x)$ be the continuous extension of $\pi_\phi^0(x)$ to an element of $B(H_\phi)$, then:
     \begin{enumerate}
-        \item $(H_\phi, \dpn{\cdot, \cdot}{\phi})$ is a Hibert space.
+        \item $H_\phi$ equipped with the unique continuous extension of $\dpn{\cdot, \cdot}{\phi}$ is a Hibert space.
         \item $(H_\phi, \pi_\phi)$ is a well-defined representation of $A$. 
-        \item $\xi_\phi = 1_A + N_\phi$ is a unit vector in $H_\phi$, and $\bracsn{\pi_\phi(x)\xi_\phi| x \in A}$ is dense in $H_\phi$. Moreover, for each $x, y \in A$,
+        \item $\xi_\phi = 1_A + N_\phi$ is a unit cyclic vector in $H_\phi$ such that for each $x, y \in A$,
         \[
         \dpn{x, y}{\phi} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(y)\xi_\phi}{H_\phi}
         \]
@@ -108,7 +108,7 @@
 
 \begin{proposition}
 \label{proposition:gns-irreducible}
-    Let $A$ be a unital $C^*$ algebra, $\phi \in S(A)$ be a state, and $(H, \pi, \xi)$ be its GNS triple, then $\phi$ is a pure state if and only if $(H, \pi)$ is irreducible. 
+    Let $A$ be a unital $C^*$-algebra, $\phi \in S(A)$ be a state, and $(H, \pi, \xi)$ be its GNS triple, then $\phi$ is a pure state if and only if $(H, \pi)$ is irreducible. 
 \end{proposition}
 \begin{proof}
     ($\Rightarrow$): Suppose that $(H, \pi)$ is reducible, then there exists a non-trivial invariant subspace $M \subset H$. Let $P \in B(H)$ be the orthogonal projection onto $M$, $\xi_1 = P\xi/\norm{P\xi}_H$, and $\xi_2 = (\xi - P\xi)/\norm{\xi - P\xi}_H$. For each $x \in A$, let
@@ -118,14 +118,21 @@
 
     then $\phi_1$ and $\phi_2$ are vector states on $A$. Since $M$ and $M^\perp$ are invariant for $\pi$, 
     \begin{align*}
-        \dpn{x, \phi}{A} &= \dpn{\pi(x)\xi, \xi}{A} = \dpn{\pi(x)P\xi, P\xi}{A} + \dpn{\pi(x)(I - P)\xi, (I - P)\xi}{A} \\
+        \dpn{x, \phi}{A} &= \dpn{\pi(x)\xi, \xi}{H} = \dpn{\pi(x)P\xi, P\xi}{H} + \dpn{\pi(x)(I - P)\xi, (I - P)\xi}{H} \\
         &= \norm{P\xi}_H^2\dpn{\pi(x)\xi_1, \xi_1}{H} + \norm{(1 - P)\xi}_H^2 \dpn{\pi(x)\xi_2, \xi_2}{H} \\
-        &= \norm{P\xi}_H^2 \dpn{x, \phi_1}{A} + \norm{(1 - P)\xi}_H^2 \dpn{x, \phi_2}{A}
+        &= \norm{P\xi}_H^2 \dpn{x, \phi_1}{A} + \norm{(I - P)\xi}_H^2 \dpn{x, \phi_2}{A}
     \end{align*}
 
-    so $\phi$ is a strict convex combination of two states, and is not pure.
+    so $\phi$ is a strict convex combination of two states. If $\phi_1 = \phi$, then for any $x, y \in A$, 
+    \begin{align*}
+    \dpn{\pi(x)\xi, \pi(y)\xi}{H} &= \norm{P\xi}_H^{-2}\dpn{\pi(x)P\xi, \pi(y)\xi}{H} \\
+    \dpn{[\pi(x) - \norm{P\xi}_H^{-2}P\pi(x)]\xi, \pi(y)\xi}{H} &= 0
+    \end{align*}
 
-    ($\Leftarrow$): Suppose that $(H, \pi)$ is irreducible. Let $t \in (0, 1)$ and $\phi_1, \phi_2 \in S(A)$ with $\phi = t\phi_1 + (1 - t)\phi_2$. Let $T: \pi(A)H \to H$ be defined by
+    so $I - \norm{P\xi}_H^{-2}P = 0$, which contradicts the fact that $P$ is a non-trivial projection. 
+    
+
+    ($\Leftarrow$): Suppose that $(H, \pi)$ is irreducible. Let $t \in (0, 1)$ and $\phi_1, \phi_2 \in S(A)$ with $\phi = t\phi_1 + (1 - t)\phi_2$. Let $T: \pi(A)\xi \to H$ be defined by
     \[
     \dpn{T\pi(x)\xi, \pi(y)\xi}{H} = \dpn{x, y}{\phi_1} = \dpn{y^*x, \phi_1}{A}
     \]
@@ -133,7 +140,7 @@
     then since $\phi_1 \le t^{-1}\phi$, $T$ extends continuously into a positive operator in $B(H)$. Moreover, for any $x, y, z \in A$, 
     \begin{align*}
     \dpn{\pi(z)T\pi(x)\xi, \pi(y)\xi}{H} &= \dpn{T\pi(x)\xi, \pi(z^*y)\xi}{H} \\
-    &= \dpn{y^*zx, \phi_1}{A} = \dpn{T\pi(z)\pi(x)\xi, \xi}{H}
+    &= \dpn{y^*zx, \phi_1}{A} = \dpn{T\pi(z)\pi(x)\xi, \pi(y)\xi}{H}
     \end{align*}
     
     so $T \in \pi(A)'$. By \hyperref[Schur's Lemma]{lemma:cstar-schur}, there exists $\lambda \in \complex$ such that $T = \lambda I$. However, since $\dpn{T\xi, \xi}{H} = \dpn{1_A, \phi_1}{A} = 1$, $\lambda = 1$, $T = I$, and $\phi_1 = \phi$. By symmetry, $\phi_2 = \phi$ as well.