C*-Algebras: Contractive Morphism

Question

Problem

Given C*-algebras $\mathcal{A}$ and $\mathcal{B}$ with $\mathbb{1}_\mathcal{A}\in\mathcal{A}$.

Consider an algebraic morphism $\pi:\mathcal{D}\subseteq\mathcal{A}\to\mathcal{B}$ with $\mathbb{1}_\mathcal{A}\in\mathcal{D}$.
(More precisely, it is a *-morphism on a *-subalgebra.)

Is it contractive then: $$\|\pi[A]\|\leq\|A\|$$

The problem is that the domain may not be closed.

Moreover, the target space may have no identity at all.

Besides, the morphism may not preserve identities.

Answer 1

I'm writing this as an answer to have a little more space to write.

What you want to prove is not true: for a $*$-homomorphism to be necessarily contractive, you need the domain to be a C$^*$-algebra.

For instance, let $\mathcal A=C[0,1]$, $\mathcal B=\mathbb C$, $\mathcal D=\{\text{polynomials}\}$, and $\pi(p)=p(2)$. Then $\pi$ is clearly a $*$-homomorphism. And it is unbounded: $$ \|x^n\|=1,\ \ \pi(x^n)=2^n,\ \ \ n\in\mathbb N. $$ Of course, this $\pi$ cannot be extended continuously to $\mathcal A$.

And this illustrates what I was mentioning in the comments: when you try to consider the inverses of polynomials, those inverses--when they exist--are not polynomials.

Answer 2

Remark

As shown in the answer above the domain must be all. Thanks alot @Martin Argerami!! :)

Proof

Suppose the domain is all $\mathcal{D}=\mathcal{A}$.

Choose a C*-subalgebra with identity: $$\mathrm{im}\pi\subseteq\mathcal{C}\subseteq\mathcal{B}:\quad1_\mathcal{C}=\pi[1_\mathcal{A}]$$

A minimal candidate is the closure of the image: $$\mathcal{C}_\text{min}=\overline{\mathrm{im}\pi}$$ and a maximal candidate is the projection onto: $$P:=\pi[\mathbb{1}_\mathcal{A}]:\quad\mathcal{C}_\text{max}=P\mathcal{B}P$$ (These really can differ!)

So one has: $$\lambda\in\rho_\mathcal{A}(A)\implies(A-\lambda1_\mathcal{A})(A-\lambda1_\mathcal{A})^{-1}=1_\mathcal{A}\text{ & vice versa}\\ \implies\pi[A-\lambda1_\mathcal{A}]\cdot\pi[(A-\lambda1_\mathcal{A})^{-1}]=\pi[1_\mathcal{A}]=1_\mathcal{C}\text{ & vice versa}\implies\lambda\in\rho_\mathcal{C}(\pi[A])$$ Note that: $\pi[A-\lambda1_\mathcal{A}]=\pi[A]-\lambda1_\mathcal{C}$

And therefore: $$\|\pi[A]\|^2=\|\pi[A^*A]\|=r_\mathcal{C}(\pi[A^*A])\leq r_\mathcal{A}(A^*A)=\|A^*A\|=\|A\|^2$$ (Note that the last both equalities turn into inequalities for involutive Banach algebras.)