If $e_j$ is an approximate unit and $\theta \in A^*$ is positive then $\|\theta e_j - \theta \| \rightarrow 0$ 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!
 A: 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. 
