In abstract algebra, we have the notion of free algebra. But does the concept of free Banach algebras also make sense?
Suppose $\mathcal{A}:=k\langle X_1, \cdots, X_n \rangle$, where $n>1$, is a finitely generated (noncommutative) free $k$-algebra, where $k$ is the field of $\mathbb{R}$ or $\mathbb{C}$, and suppose a norm $\|\cdot\|$ makes $\mathcal{A}$ a normed algebra such that $\|a\cdot_\mathcal{A} b\|\leq \|a\|\|b\|$, for all $a,b\in \mathcal{A}$. Then, is the completion $\hat{\mathcal{A}}$ w.r.t this norm a free Banach algebra? The noncommutative multiplication in $\hat{\mathcal{A}}$ is defined as the natural extension of the multiplications in $\mathcal{A}$: For $a'=\lim a_n$ and $b'=\lim b_n$ in $\hat{\mathcal{A}}$, where $(a_n), (b_n)\subset \mathcal{A}$, $$a' \cdot_{\hat{\mathcal{A}}} b':=\lim_{n\to \infty} (a_n\cdot_\mathcal{A} b_n). $$
If the above construction does make sense, are there well-known concrete examples of (noncommutative) free Banach algebras out there? Thanks in advance.
Yes, just take polynomial functions on $[0,1]$ with $\|\ast\|_\infty$ as a norm, then $A \simeq \mathbb{R} [X]$, and $\hat{A} = C^0 ([0,1])$. The only issue here (and always) is that you won't find a finitely generated algebra. Probably use Baire to show it.