I found the following exercise in my lecture notes: Let $\mathcal{A}$ be a $C^*$-algebra. Then every multiplicative functional is an extremal point of $K_1^{\mathcal{A}'}(0)$.
Other than writing $m = t m_1 + (1-t) m_2$ for a multiplicative functional $m$, $t \in (0,1)$ and $m_1,m_2 \in K_1^{\mathcal{A}'}(0)$ and trying to show that $m_1 = m = m_2$, I got nothing. I thought (assuming commutativity) of using the Gelfand-Neimark theorem to identify my $C^*$-algebra with the space of continuous functions. But I was not able to deduce anything from it.
Question: What would be a good starting point to tackle the problem?
Thanks in advance!
Assume first that $A$ is unital. Then $$1 = m(\mathbf{1}) = tm_1(\mathbf{1}) + (1-t)m_2(\mathbf{1})$$ and $|m_1(\mathbf{1})|, |m_2(\mathbf{1})|\le 1$ because $\|\mathbf{1}\| = 1$ and $\|m_1\|, \|m_2\| \le 1$ by assumption. It follows that $m_1(\mathbf{1}) = m_2(\mathbf{1}) = 1$ and $\|m_1\| = \|m_2\| = 1$.
By proposition 1.12.2 of this introduction to $C^*$-algebras, it follows that $m_1, m_2$ are positive linear functionals. Let $U = \max(\frac{1}{t}, \frac{1}{1-t})$. If $a\in A, a\ge 0$ then $m_1(a), m_2(a)\ge 0$ so from the equation $m(a) = tm_1(a) + (1-t)m_2(a)$ we conclude that $m_1(a), m_2(a) \le U m(a)$.
By lemma 1.12.3 part 1 of same introduction, for any $a,b\in A$ $$|m_1(b^* a)|^2 \le m_1(a^* a) m_1(b^* b)$$ So now let $a\in A$ such that $m(a) = 0$ and take $b=\mathbf{1}$ (so $m_1(b^* b) = m_1(\mathbf{1}) = 1$). Then $$|m_1(a)|^2\le m_1(a^* a) \le U m(a^* a) = U m(a^*) m(a) = 0$$ where $m(a^* a) = m(a^*) m(a)$ by the multiplicativity of $m$.
This proves that $\operatorname{Ker}(m) \subseteq \operatorname{Ker}(m_1)$. But these are linear functionals, so $\operatorname{Ker}(m)$ is a linear subspace of $A$ of co-dimension $1$, and $m_1\ne 0$ (because $m_1(\mathbf{1}) = 1$). It follows that $m_1 = m$.
If $A$ is not unital, you can adjoin a unit to $A$ (see here, section 2.4.) I haven't checked the detail, but I believe that by working in the algebra after adjoining a unit to $A$ you can get the result for non-unital $A$ as well.