Is the supremum norm the only $ C^{*} $-norm on $ {C_{c}}(X) $, equipped with the usual pointwise operations?

Question

Let $ X $ be a locally compact Hausdorff space. Then $ {C_{c}}(X) $ is a commutative $ * $-algebra with respect to addition, multiplication, scalar multiplication and conjugation (all pointwise operations).

Question. If $ \| \cdot \|: {C_{c}}(X) \to \Bbb{R}_{\geq 0} $ is a (not-necessarily-complete) $ C^{*} $-norm on this $ * $-algebra, then is it true that $ \| \cdot \| $ is actually the supremum norm on $ {C_{c}}(X) $? If so, then $ {C_{0}}(X) $ is the only $ C^{*} $-completion of this $ * $-algebra.

This question seemed trivial at first sight, but after a while, I realized that it was not at all.

Thanks for your help!

Answer 1

Let $\|\cdot\|$ be some C$^*$-norm on $C_c(X)$. Write $A=(C_0 (X),\|\cdot\|_\infty) $, $B=\overline {C_c(X),\|\cdot\|) }$.

Fix $f_0\in C_c(X)$. Put $A_0=C^*(f_0)\subset A$, and $\pi:A_0\to B$ the identity map (this works because $A_0\subset C_c (X)\subset B $; since we use the supremum norm to generate it, every element has support inside that of $f_0$).

As $\pi$ is a $*$-monomorphism it is isometric, and thus $\|f_0\|=\|f_0\|_\infty$.