[proofplan] We first extend the state $\rho$ from the closed linear subspace $B\subset A$ to a bounded complex linear functional on all of $A$ by the complex [Hahn-Banach theorem](/theorems/879), preserving its norm. Since $A$ and $B$ have the same unit, the extension still takes the value $1$ at the unit. The remaining point is to prove a standard characterization: a complex linear functional $\phi$ on a unital C*-algebra satisfying $\|\phi\|=\phi(1)=1$ is positive. Applying that characterization to the Hahn-Banach extension gives a positive normalized extension, hence a state. [/proofplan]

[step:Extend the state by complex Hahn-Banach with unchanged norm] Because $\rho$ is a state on the unital C*-algebra $B$, it is positive and satisfies $\rho(1_A)=1$. By [citetheorem:8557], applied to the positive linear functional $\rho:B\to\mathbb C$, its operator norm is \begin{align*} \|\rho\|_{B^*}=\rho(1_A)=1. \end{align*} Regard $B$ as a complex linear subspace of the normed complex [vector space](/page/Vector%20Space) $A$, with the norm inherited from $A$. By the complex Hahn-Banach theorem for normed complex vector spaces (citing a result not yet in the wiki: complex Hahn-Banach theorem for normed complex vector spaces), there exists a complex linear functional \begin{align*} \widetilde{\rho}:A\to\mathbb C \end{align*} such that $\widetilde{\rho}|_B=\rho$ and \begin{align*} \|\widetilde{\rho}\|_{A^*}=\|\rho\|_{B^*}=1. \end{align*} Since $1_A\in B$ and the units of $A$ and $B$ agree, the extension property gives \begin{align*} \widetilde{\rho}(1_A)=\rho(1_A)=1. \end{align*} [/step]

[step:Prove that every norm-one unital functional is positive][claim:Norm-one unital functionals on unital C*-algebras are positive] Let $D$ be a unital C*-algebra with unit $1_D$, and let $\phi:D\to\mathbb C$ be a complex linear functional satisfying \begin{align*} \|\phi\|_{D^*}=1 \end{align*} and \begin{align*} \phi(1_D)=1. \end{align*} Then $\phi(x)\ge 0$ for every positive element $x\in D$. [/claim] [proof] Let $h\in D$ be self-adjoint and assume $\|h\|_D\le 1$. For each $t\in\mathbb R$, the C*-identity and submultiplicativity give \begin{align*} \|1_D+ith\|_D^2=\|(1_D+ith)^*(1_D+ith)\|_D. \end{align*} Since $h^*=h$ and $t\in\mathbb R$, \begin{align*} (1_D+ith)^*(1_D+ith)=(1_D-ith)(1_D+ith)=1_D+t^2h^2. \end{align*} Therefore \begin{align*} \|1_D+ith\|_D^2=\|1_D+t^2h^2\|_D\le \|1_D\|_D+t^2\|h^2\|_D. \end{align*} Using $\|1_D\|_D=1$ and $\|h^2\|_D\le \|h\|_D^2\le 1$, we obtain \begin{align*} \|1_D+ith\|_D\le (1+t^2)^{1/2}. \end{align*} Write $\phi(h)=\alpha+i\beta$, where $\alpha,\beta\in\mathbb R$. Since $\|\phi\|_{D^*}=1$ and $\phi(1_D)=1$, \begin{align*} |1+it\phi(h)|=|\phi(1_D+ith)|\le \|1_D+ith\|_D\le (1+t^2)^{1/2}. \end{align*} Squaring both sides gives \begin{align*} (1-t\beta)^2+t^2\alpha^2\le 1+t^2. \end{align*} Equivalently, \begin{align*} -2t\beta+t^2(\alpha^2+\beta^2)\le t^2. \end{align*} For $t>0$, dividing by $t$ gives \begin{align*} -2\beta+t(\alpha^2+\beta^2)\le t. \end{align*} Letting $t\to 0$ through positive [real numbers](/page/Real%20Numbers) yields $\beta\ge 0$. For $t<0$, dividing by $t$ reverses the inequality and gives \begin{align*} -2\beta+t(\alpha^2+\beta^2)\ge t. \end{align*} Letting $t\to 0$ through negative real numbers yields $\beta\le 0$. Hence $\beta=0$, so $\phi(h)\in\mathbb R$ for every self-adjoint $h\in D$ with $\|h\|_D\le 1$. If $k\in D$ is self-adjoint and nonzero, applying this to $h=k/\|k\|_D$ gives $\phi(k)\in\mathbb R$; the case $k=0$ is immediate. Thus $\phi$ is real-valued on all self-adjoint elements of $D$. Now let $x\in D$ be positive. If $x=0$, then $\phi(x)=0$. Assume $x\ne 0$, and define \begin{align*} h:=\frac{x}{\|x\|_D}. \end{align*} Then $h$ is positive and $\|h\|_D=1$. By the standard C*-order norm fact for positive contractions, $0\le h\le 1_D$, so $1_D-h$ is positive and \begin{align*} \|1_D-h\|_D\le 1. \end{align*} The element $1_D-h$ is self-adjoint, so $\phi(1_D-h)\in\mathbb R$. Moreover, \begin{align*} |\phi(1_D-h)|\le \|\phi\|_{D^*}\|1_D-h\|_D\le 1. \end{align*} Since $\phi(1_D)=1$, \begin{align*} \phi(1_D-h)=1-\phi(h). \end{align*} Thus \begin{align*} -1\le 1-\phi(h)\le 1. \end{align*} In particular, $\phi(h)\ge 0$. Multiplying by the positive scalar $\|x\|_D$ gives \begin{align*} \phi(x)=\|x\|_D\phi(h)\ge 0. \end{align*} Therefore $\phi$ is positive. [/proof][/step]

[guided]We prove the criterion from the ground up because this is the point where positivity enters. Let $D$ be a unital C*-algebra with unit $1_D$, and let $\phi:D\to\mathbb C$ be complex linear with $\|\phi\|_{D^*}=1$ and $\phi(1_D)=1$. The goal is to show that $\phi(x)\ge 0$ whenever $x\in D$ is positive. First we show that $\phi$ takes real values on self-adjoint elements. Let $h\in D$ be self-adjoint and satisfy $\|h\|_D\le 1$. For a real parameter $t\in\mathbb R$, we estimate $1_D+ith$. The reason for this particular element is that the C*-identity turns its norm into a norm involving $h^2$, while applying $\phi$ to it exposes the imaginary part of $\phi(h)$. By the C*-identity, \begin{align*} \|1_D+ith\|_D^2=\|(1_D+ith)^*(1_D+ith)\|_D. \end{align*} Because $h^*=h$ and $t$ is real, \begin{align*} (1_D+ith)^*(1_D+ith)=(1_D-ith)(1_D+ith)=1_D+t^2h^2. \end{align*} Therefore, using the triangle inequality and submultiplicativity of the C*-norm, \begin{align*} \|1_D+ith\|_D^2=\|1_D+t^2h^2\|_D\le \|1_D\|_D+t^2\|h^2\|_D. \end{align*} Since $D$ is unital, $\|1_D\|_D=1$, and since $\|h\|_D\le 1$, \begin{align*} \|h^2\|_D\le \|h\|_D^2\le 1. \end{align*} Thus \begin{align*} \|1_D+ith\|_D\le (1+t^2)^{1/2}. \end{align*} Now write $\phi(h)=\alpha+i\beta$ with $\alpha,\beta\in\mathbb R$. The boundedness of $\phi$ gives \begin{align*} |1+it\phi(h)|=|\phi(1_D+ith)|\le \|\phi\|_{D^*}\|1_D+ith\|_D. \end{align*} Using $\|\phi\|_{D^*}=1$, this becomes \begin{align*} |1+it\phi(h)|\le (1+t^2)^{1/2}. \end{align*} Since \begin{align*} 1+it(\alpha+i\beta)=1-t\beta+it\alpha, \end{align*} squaring absolute values gives \begin{align*} (1-t\beta)^2+t^2\alpha^2\le 1+t^2. \end{align*} Expanding and cancelling $1$ from both sides yields \begin{align*} -2t\beta+t^2(\alpha^2+\beta^2)\le t^2. \end{align*} For positive $t$, division by $t$ preserves the inequality: \begin{align*} -2\beta+t(\alpha^2+\beta^2)\le t. \end{align*} Letting $t\to 0$ through positive real numbers gives $\beta\ge 0$. For negative $t$, division by $t$ reverses the inequality: \begin{align*} -2\beta+t(\alpha^2+\beta^2)\ge t. \end{align*} Letting $t\to 0$ through negative real numbers gives $\beta\le 0$. Hence $\beta=0$, and therefore $\phi(h)\in\mathbb R$. This proves reality for self-adjoint contractions. If $k\in D$ is any nonzero self-adjoint element, then $k/\|k\|_D$ is a self-adjoint contraction, so $\phi(k/\|k\|_D)\in\mathbb R$. By complex linearity, \begin{align*} \phi(k)=\|k\|_D\phi\left(\frac{k}{\|k\|_D}\right)\in\mathbb R. \end{align*} If $k=0$, then $\phi(k)=0$. Thus $\phi$ is real-valued on every self-adjoint element. We now prove positivity. Let $x\in D$ be positive. If $x=0$, then $\phi(x)=0$. Suppose $x\ne 0$, and define the normalized positive element \begin{align*} h:=\frac{x}{\|x\|_D}. \end{align*} Then $h$ is positive and $\|h\|_D=1$. The standard C*-order norm fact for positive contractions says that $0\le h\le 1_D$, hence $1_D-h$ is positive and \begin{align*} \|1_D-h\|_D\le 1. \end{align*} Since $1_D-h$ is self-adjoint, the reality result already proved implies \begin{align*} \phi(1_D-h)\in\mathbb R. \end{align*} The norm estimate for $\phi$ gives \begin{align*} |\phi(1_D-h)|\le \|\phi\|_{D^*}\|1_D-h\|_D\le 1. \end{align*} Using $\phi(1_D)=1$, we rewrite the scalar on the left as \begin{align*} \phi(1_D-h)=1-\phi(h). \end{align*} Because this scalar is real and has absolute value at most $1$, \begin{align*} -1\le 1-\phi(h)\le 1. \end{align*} The left inequality gives $\phi(h)\ge 0$. Finally, $x=\|x\|_D h$ and $\|x\|_D>0$, so \begin{align*} \phi(x)=\|x\|_D\phi(h)\ge 0. \end{align*} Thus $\phi$ is positive on every positive element of $D$.[/guided]

[step:Apply the positivity criterion to the Hahn-Banach extension] Apply the claim with $D=A$ and $\phi=\widetilde{\rho}$. The previous step verifies the two hypotheses: \begin{align*} \|\widetilde{\rho}\|_{A^*}=1 \end{align*} and \begin{align*} \widetilde{\rho}(1_A)=1. \end{align*} Hence $\widetilde{\rho}(a)\ge 0$ for every positive element $a\in A$. Therefore $\widetilde{\rho}:A\to\mathbb C$ is a positive linear functional with $\widetilde{\rho}(1_A)=1$, so it is a state on $A$. Since $\widetilde{\rho}|_B=\rho$ by construction, this state extends $\rho$, as required. [/step]