I am currently working through Murphy's book C$^*$-Algebras and Operator Theory, and in the beginning of chapter 5 he mentions the following in passing:

If $(H,\varphi)$ is a non-degenerate representation of a C$^*$-algebra $A$, and $\{u_\lambda\}$ is an approximate identity for $A$, then $\{\varphi(u_\lambda)\}$ is an approximate unit for $\varphi(A)$, so the net $\{\varphi(u_\lambda)\}$ converges strongly to $\text{id}_H$.

Recall that a representation $(H,\varphi)$ of a C$^*$-algebra is non-degenerate if $\overline{\varphi(A)H}=H$, or equivalently for each nonzero $x\in H$ there is some $a\in A$ such that $\varphi(a)x\neq0$.

I'm having trouble seeing why this is true. By virtue of $\varphi$ being a $*$-homomorphism, we know that $\{\varphi(u_\lambda)\}$ is an increasing net (since $\varphi$ is positive) in the closed unit ball of $\varphi(A)$ (since $\varphi$ is norm-decreasing). Furthermore, for all $a\in A$ we have $$\|\varphi(u_\lambda a-a)\|\leq\|u_\lambda a-a\|$$ which should converge to zero. However, this doesn't require $(H,\varphi)$ be non-degenerate, and it seems like it should be necessary.

Could someone please help me see some error in my judgement, or possibly provide a correct proof of the above statement? Any help would be greatly appreciated.


1 Answer 1


Since the representation is non-degenerate, $\{\varphi(a)x:\ a\in A,\ x\in H\}$ is dense in $H$.

For such $\varphi(a)x\in H$, $$ \|\varphi(u_\lambda)\varphi(a)x-\varphi(a)x\|=\|\varphi(u_\lambda a-a)x\| \leq\|\varphi(u_\lambda a -a)\|\leq\|u_\lambda a - a\|\to0. $$ So $\varphi(u_\lambda)\to I$ strongly on a dense subset, and now an approximation argument shows that $\varphi(u_\lambda)\to I$ strongly on $H$.

  • $\begingroup$ Oh ok, so the non-degeneracy is just needed to show that $\{\varphi(u_\lambda)\}$ converges strongly to $I$. Thanks! $\endgroup$
    – Aweygan
    Apr 28, 2017 at 22:40
  • $\begingroup$ Glad I could help! $\endgroup$ Apr 28, 2017 at 22:45

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