= Solution
In a unital <C-star algebra> $A$, an element is <Hermitian element of a C-star algebra>[Hermitian] when $x=x^*$, <Unitary element of a C-star algebra>[unitary] when $x^*x=xx^*=1$, and <Normal element of a C-star algebra>[normal] when $x^*x=xx^*$. It is a <positive element of a C-star algebra> when $x=y^*y$ for some $y$, equivalently when it is Hermitian and its spectrum lies in $[0,\infty)$.
Suppose first that $\|\tau\|=\tau(1)$, and normalize so that both equal one. If $x=x^*$, the C-star identity gives
$$
\|x+it1\|^2
=\|(x-it1)(x+it1)\|
=\|x^2+t^21\|
\leq\|x\|^2+t^2.
$$
Writing $\tau(x)=a+ib$, we obtain
$$
a^2+(b+t)^2=|\tau(x+it1)|^2\leq\|x\|^2+t^2
$$
for every real $t$, which forces $b=0$. Decomposing an arbitrary $x$ into its Hermitian real and imaginary parts now gives
$$
\boxed{\tau(x^*)=\overline{\tau(x)}}.
$$
Let $x=x^*$. Every <character of an algebra> on the commutative unital Banach algebra generated by $x$ has norm and value at one equal to one, so the preceding argument makes its value on $x$ real. The character description of the <spectrum of an element> therefore gives $\sigma(x)\subset\mathbb R$. Iterating the C-star identity,
$$
\|x\|=\|x^{2^n}\|^{1/2^n}\longrightarrow r(x),
$$
so $\|x\|=r(x)$. If $c=\|x\|$, then $x+c1$ has nonnegative spectrum and is positive, as is $c1$; hence
$$
x=(x+c1)-c1
$$
is a difference of positive elements.
If $0\leq x$ and $\|x\|\leq1$, then spectral translation gives
$$
\sigma(1-x)=1-\sigma(x)\subseteq[0,1].
$$
Thus $1-x$ is positive and, by the norm--spectral-radius equality for Hermitian elements, $\|1-x\|\leq1$.
Return to a norm-one $\tau$ with $\tau(1)=1$. For $0\leq x\leq1$, the preceding paragraph and reality on Hermitian elements give
$$
\tau(x)=1-\tau(1-x)\geq1-|\tau(1-x)|\geq0.
$$
Scaling proves that $\tau$ is a <positive functional on a C-star algebra>. Every character has norm and value at one equal to one, so every character is positive. On $C[0,1]$, the functional
$$
f\longmapsto\int_0^1f(t)\,dt
$$
is positive but is not multiplicative, and hence is not a character.
Conversely, let $\tau$ be positive. Writing a Hermitian element as a difference of positive elements shows that $\tau$ is real on Hermitian elements. Positivity of
$$
\tau((x\lambda+y)^*(x\lambda+y))
$$
for every $\lambda\in\mathbb C$ says that the associated quadratic polynomial is nonnegative. Minimizing it in $\lambda$ gives the Cauchy--Schwarz inequality
$$
\boxed{|\tau(y^*x)|^2\leq\tau(x^*x)\tau(y^*y)}.
$$
Taking $y=1$ yields $|\tau(x)|^2\leq\tau(x^*x)\tau(1)$. Since $0\leq x^*x\leq\|x\|^21$, positivity gives
$$
|\tau(x)|\leq\tau(1)\|x\|.
$$
Together with $\tau(1)\leq\|\tau\|$, this proves $\|\tau\|=\tau(1)$.
Convex combinations preserve positivity and value one at the identity, so the <state on a C-star algebra>[state space] $S(A)$ is convex. If $x$ is normal, the unital C-star subalgebra $C^*(1,x)$ is commutative and its <Gelfand transform> identifies it with $C(\sigma(x))$. Choose $\lambda\in\sigma(x)$ with $|\lambda|=\|x\|$. Evaluation at $\lambda$ is a state taking $x$ to $\lambda$. Its norm-preserving <Hahn-Banach theorem>[Hahn--Banach extension] to $A$ still takes $1$ to one, so the norm criterion makes the extension a state $\tau$ satisfying $|\tau(x)|=\|x\|$.
The state space is nonempty, convex, and weak-star compact by the <Banach-Alaoglu theorem>. The <Krein-Milman theorem> gives an extreme point, so a <pure state on a C-star algebra> exists. For positive $x$, the preceding norm-attainment result makes
$$
F=\{\tau\in S(A):\tau(x)=\|x\|\}
$$
a nonempty weak-star compact convex set. It is a face: if a convex combination has the maximal possible value on $x$, each summand does. An extreme point of $F$ exists by Krein--Milman and, because $F$ is a face, is extreme in $S(A)$. It is the required pure state.
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