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Let $A$ be a C$^*$-algebra, $\mathbb{H}$ a Hilbert space, and $\pi:A \to B(\mathbb{H})$ a faithful $*$-representation, for which $B(\mathbb{H})$ is the space of bounded linear operators on $\mathbb{H}$.

My questions:

i) Will $\pi$ be isometric?

ii) Will it's image be closed?

iii) What happends if we remove faithfulness?

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    $\begingroup$ i) injective $C^*$-morphisms are isometric. As a consequence for ii) the image is going to be closed. For iii) if you remove faithfullness you will definitely get a failure of isometric, but the image of a $*$-morphism is always closed. $\endgroup$ – s.harp Aug 12 at 18:18
  • $\begingroup$ Near-duplicate: math.stackexchange.com/questions/1400305/… $\endgroup$ – Eric Wofsey Aug 12 at 18:23
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i) Yes.

ii) Yes.

iii) It won't be isometric, but its image will be closed.

This stems from the following result:

Let $A$ and $B$ be $C^*$-algebras, and let $\varphi:A\to B$ be a $*$-homomorphism. If $\varphi$ is injective, then it is an isometry.

A proof of this can be found in most introductory books on $C^*$-algebras, for example, in chapter 1 of Davidson's $C^*$-Algebras by Example, or in chapter 3 of Murphy's $C^*$-Algebras and Operator Theory. This gives i).

The positive answer for ii) follows from the positive answer to i) and from a standard result in functional analysis:

If $X$ and $Y$ are Banach spaces and $T:X\to Y$ is an isometric linear map, then $T(X)$ is closed.

To answer iii), we use a corollary of the first result:

Let $A$ and $B$ be $C^*$-algebras, and let $\varphi:A\to B$ be a (not necessarily injective) $*$-homomorphism. Then $\varphi(A)$ is a $C^*$-subalgebra of $B$ (i.e., its image is closed).

This comes from the fact that the image of $\varphi:A\to B$ is the same as the image of the induced $*$-homomorphism $\tilde\varphi:A/\ker(\varphi)\to B$, which is injective, hence isometric, hence has closed range.

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