Given $\cal{A},\cal{B}$ two dense $\*$-algebras of two $C^\*$-algebras $A$ and $B$ respectively, together with a $\*$-algebra homomorphism $f:\cal{A} \to \cal{B}$, is it clear that $f$ extends to a bounded linear operator $f:A \to B$?



Extending maps from dense $*$-algebras of $C^*$-algebras
I believe this is a counter-example.


Let $\newcommand{\mc}{\mathcal}\mc A$ be the algebra of complex polynomials restricted to $[0,1]$, with closure $A=C[0,1]$. Let $X\in\mc A$ be the coordinate function, $X(t)=t$ for $t\in[0,1]$, so $X$ generates $\mc A$ as a $\*$-algebra.


Let $H$ be a Hilbert space and $x\in\mc B(H)$ be any non-zero self-adjoint operator. Let $\mc B$ be the $\*$-algebra generated by $x$ and let $B$ be the closure, a $C^\*$-algebra.


Let $f:\mc A\rightarrow\mc B$ be the unique $\*$-homomorphism with $f(X)=x$. Suppose that $f$ extends by continuity to $A$, say $\pi:A\rightarrow B$ with $\pi$ bounded. By continuity, $\pi$ is a $\*$-homomorphism, and so is contractive. But then $\|x\| = \|\pi(X)\| \leq \|X\|=1$. As not every self-adjoint operator is a contraction, this provides the required contradiction.


