# A faithful $*$-representation of a C$^*$-algebra

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?

• 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. Aug 12 '19 at 18:18
• Near-duplicate: math.stackexchange.com/questions/1400305/… Aug 12 '19 at 18:23

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.