Here is a quotation of a book "C*-algebras and Finite-Dimensional Approximations" P275
Let $A$ be a C*-algebra. Suppose $I\triangleleft A$ be a closed ideal. If the representation $I\hookrightarrow B(H)$ is a nondegenerate representation, then the map $I\hookrightarrow B(H)$ can be uniquely to a *-homomorphism $\pi: A\rightarrow B(H)$ by defining $$\pi(a)=lim ae_{n}$$ where $\{e_{n}\}\subset I$ is any approximate unit and the limit is taken in the strong operator topology (uniqueness of $\pi$ is again due to the fact that $I\subset B(H)$ is nondegenerate).
I have two questions about the "extension" above:
Why does the author take the limit in the strong operator topology, not in norm topology? I think if the author define the limit in norm topology, then it is easy to see that this homomorphism is an extension.
How to prove the uniqueness of $\pi$ here? Actually, I do not know why should we verify the uniqueness.
The limit may not exist in the norm topology (for instance if $I = K(H)$ and $\{e_n\}$ is the usual collection of finite rank projections converging SOT to the identity). However, the SOT limit exists, since $\{e_n\}$ is SOT-convergent (being an increasing sequence of self-adjoint operators that are bounded above)
As for uniqueness, suppose $\varphi$ is any other extension, we need to check that $$ \pi(a)y = \varphi(a)y $$ for any $y$ in a dense subset of $H$. Since $I \subset B(H)$ is nondegenerate, the set $$ \{x\xi : x\in I, \xi \in H\} $$ is dense in $H$, and $$ \pi(a)x\xi = \pi(ax)\xi = ax\xi = \varphi(ax)\xi = \varphi(a)x\xi $$