Let $A$ be a $C^*$-algebra and $e_j$ where $j \in \mathbb{N}$ be an approximate unit.
Let $\theta \in A^*$ be positive and such that for $a \in A$ we write $\theta a \in A^*$ for the functional $(\theta a)(x)=\theta (ax)$ where $x \in A$.
Prove that $\|\theta e_j - \theta\| \rightarrow 0$
I don't know how to start proving the above. Any help is appreciated!
Consider the GNS representation of $\theta$. That is, we assume $$ \theta(a)=\langle \pi(a)\xi,\xi\rangle,\ \ \ \ a\in A, $$ for some Hilbert space $H$, a representation $\pi: A\to B(H)$, and a fixed $\xi\in H$ with $\pi(A)\xi$ dense. Since $\{e_j\}$ is an approximate unit, it is standard that $\pi(e_j)\xrightarrow{\rm sot} I$. So, for $a\in A$, \begin{align} |\theta(ae_j-a)|&=|\langle\pi(ae_j-a)\xi,\xi\rangle|=|\langle(\pi(e_j)-I)\xi,\pi(a)^*\xi\rangle|\\ \ \\ &\leq \|\pi(a)^*\xi\|\,\|(\pi(e_j)-I)\xi\|\\ \ \\ &\leq\|a\|\,\|(\pi(e_j)-I)\xi)\|. \end{align} It follows that $$ \|\theta e_j-\theta\|\leq\|(\pi(e_j)-I)\xi\|. $$
Edit: a proof that $e_j\xrightarrow{\rm sot}I$.
Let $a\in A$. Using that $0\leq e_j\leq I$ (so in particular $e_j^2\leq e_j$), \begin{align} \|(e_j-I)a\xi\|^2&=\|e_ja\xi-a\xi\|^2=\|e_ja\xi\|^2+\|a\xi\|^2-2\text{Re}\,\langle e_ja\xi,a\xi\rangle \\ \ \\ &=\langle a^*e_j^2a\xi,\xi\rangle+\langle a^*a\xi,\xi\rangle-2\langle a^*e_ja\xi,\xi\rangle\\ \ \\ &\leq\langle a^*e_ja\xi,\xi\rangle+\langle a^*a\xi,\xi\rangle-2\langle a^*e_ja\xi,\xi\rangle\\ \ \\ &=\langle a^*a\xi,\xi\rangle-\langle a^*e_ja\xi,\xi\rangle\\ \ \\ &=\theta(a^*a-a^*e_ja)\leq \|a^*a-a^*e_ja\|\\ \ \\ &\leq\|a\|\,\|a-e_ja\|\to0. \end{align} Now, for arbitrary $\eta\in H$, fix $\varepsilon>0$. Then there exists $a\in A$ with $\|\eta-a\xi\|<\varepsilon$. So \begin{align} \|(e_j-I)\eta\|&\leq\|(e_j-I)a\xi\|+\|(e_j-I)(\eta-a\xi)\|\\ \ \\ &\leq \|(e_j-I)a\xi\|+2\|\eta-a\xi\|\\ \ \\ &\leq\|(e_j-I)a\xi\|+2\varepsilon. \end{align} Thus $$ \limsup_j\|(e_j-I)\eta\|\leq2\varepsilon. $$ As $\varepsilon$ was arbitrary, the limit does exist and equals zero.