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In the proof of Lemma A.4 of this document, the author proves that a C$^*$-algebra is simple by showing that every one of its representations is faithful. They proceed by taking a representation and show it is injective on a (fixed) dense $^*$-subalgebra. Then they claim that the restriction of the representation to this subalgebra is faithful (obvious) and that the extension back to the original algebra is faithful (valid conclusion?).

Perusing other questions on here it would appear that there is a problem with the final claim, if we just look at extending continuous injective functions then you can't just conclude that the extension is injective, and if we look at extending injective $^*$-homomorphisms on dense $^*$-subalgebras, then there are still problems in the general case.

I suppose that my question would be the following; what extra conditions may we impose on an injective $^*$-homomorphism $\varphi:\mathcal{D}\rightarrow B$ between a dense $^*$-subalgebra $\mathcal{D}$ of a C$^*$-algebra $A$, and another C$^*$-algebra $B$ so that it's extension to $A$ is injective?

The only thing which springs to mind is that if we insist that $\varphi$ is an isometry, then its extension will also be an isometry, hence injective.

user505379
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First, as you say it is not true that a faithful representation on a dense subalgebra extends to a faithful representation. For an example, let $A=C[0,1]$, $D$ the polynomials, and let $\pi:D\to \ell^\infty$ be given by $$ \pi(p)=(p(1),p(1/2),p(1/3),\ldots). $$ Then $\pi$ is faithful, but its (unique) extension to $C[0,1]$ is not (as there exist many functions $f$ with $f(1/n)=0$ for all $n\in\mathbb N$).

As for a sufficient condition, one possibility is to require $D$ to have the property that for any positive $a\in A$ there exists positive $d\in D$ with $d\leq a$. Then $\pi(a)\geq \pi(d)$, and so $\pi(a)$ cannot be zero for any positive $a\in A$.

Martin Argerami
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  • The polynomials aren't dense in $C_0(\mathbb R)$, in fact, the only polynomial in $C_0(\mathbb R)$ is the constant $0$ polynomial. – Aweygan Aug 09 '18 at 22:49
  • But this can be modified to $A=C([0,1])$, $D$ the polynomials, $\pi: D\to\ell^\infty$ by $\pi(p)=(p(1),p(\frac{1}{2}),p(\frac{1}{3}),\ldots)$. – Aweygan Aug 09 '18 at 23:01
  • You are right, I didn't use $[0,1]$ to go for the easy sequence, but it doesn't work. I'll edit. – Martin Argerami Aug 10 '18 at 02:31