Consider a $C^*$-algebra $A$ and a continuous linear functional $\varphi:A\to\mathbb{C}$ with adjoint given by $\varphi^*(a):=\overline{\varphi(a^*)}$. We know that each positive (cont. lin.) functional is self-adjoint.
How would one show that each $\varphi\in A^*$ can be written as a linear combination of two self-adjoint members of $A^*$?
My attempt:
I thought I'd try with $\varphi_1(a+ib):=\varphi(a)$ and $\varphi_2(a+ib):=\varphi(b)$, though I am unsure whether $\varphi_1,\varphi_2$ are indeed self-adjoint. I manage to conclude that $\varphi_1^*(a+ib)=\overline{\varphi(a)}$ and $\varphi_1^*(a+ib)=\overline{\varphi(b)}$, but am unsure if concluding those equal to $\varphi(a),\varphi(b)$ would be legal. Note that both $a$ and $b$ are self-adjoint.
$$ \begin{align} \varphi^{**}(a)=\overline{\varphi^*(a^*)}&=\overline{\overline{\varphi(a^{**})}}=\varphi(a)\\ (\varphi+\psi)^*(a)=\overline{(\varphi+\psi)(a^*)}&= \overline{\varphi(a^*)+\psi(a^*)}=\overline{\varphi(a^*)}+\overline{\psi(a^*)} =\varphi^*(a)+\psi^*(a)\\ (\lambda \varphi)^*(a)=\overline{\lambda\varphi(a^*)}&= \bar{\lambda}\overline{\varphi(a^*)}=\bar{\lambda}\varphi^*(a) \end{align} $$ Any time you have a vector space $V$ over $\Bbb{C}$ with a complex linear involution $i\colon V\to V$ with the properties $$ \begin{align} i(i(v))&=v\\ i(v+w)&=i(v)+i(w)\\ i(\lambda v)&=\bar\lambda i(v) \end{align} $$ the self-adjoint elements are $v\in V$ such that $i(v)=v$ and any $v\in V$ has a decomposition $v=\Re(v)+\sqrt{-1}\Im(v)$ where $$ \begin{split} \Re(v)=\frac{v+i(v)}{2}\\ \Im(v)=\frac{v-i(v)}{2\sqrt{-1}} \end{split} $$ and $\Re(v),\Im(v)$ are self-adjoint.