Let $\mathbb{C}\langle x,y\rangle$ be the group ring of the complex numbers over the free group in $x,y$. Let $len : \langle x,y \rangle \rightarrow \mathbb{N}$ denote the standard word norm and let $\varphi=exp \circ len: \langle x, y \rangle \rightarrow [1,\infty)$. Define the following norm for $\displaystyle \alpha=\sum_{g \in \langle x,y\rangle} a_{g}g \in \mathbb{C}\langle x,y \rangle$:

$$\|\alpha\|=\sum_{g \in \langle x, y \rangle} |a_{g}|\varphi(g)$$

It's not hard to show that $\|\cdot\|$ is indeed a norm and in fact for all $\alpha, \beta \in \mathbb{C}\langle x, y \rangle$ we have $\|\alpha\beta\| \le \|\alpha\|\cdot\|\beta\|$. The natural involution $\displaystyle ^{*}:\sum_{g \in \langle x, y \rangle}a_{g}g \mapsto \sum_{g \in \langle x, y \rangle} \overline{a_{g}}g^{-1}$ has the properties that would make $(\mathbb{C}\langle x, y \rangle,\|\cdot\|,^{*})$ into a Banach *-algebra, including $\|\alpha\|=\|\alpha^{*}\|$. My question is this:

Is $\mathbb{C}\langle x, y \rangle$ complete with respect to $\|\cdot\|$?

I stumbled upon this while working on an undergrad research project but haven't had any functional analysis, so I'm not sure how to go about proving/disproving this. Any help/references would be greatly appreciated.

  • $\begingroup$ When you write $r_g$ do you mean $a_g$? $\endgroup$ Jun 7, 2012 at 20:19
  • $\begingroup$ Yup, thanks for catching that and thanks for your answer! $\endgroup$ Jun 7, 2012 at 20:30

1 Answer 1


There are no countable-dimensional Banach spaces. This is a corollary of the Baire category theorem: if $e_1, e_2, ...$ were a basis for a Banach space $B$, then $B$ would be the countable union of the sets $\text{span}(e_1), \text{span}(e_1, e_2), ...$, all of which are nowhere dense, which contradicts BCT3 (in Wikipedia's terminology).

There is an obvious and more naive argument which doesn't work: if I had $e_1, e_2, ...$ as above, normalized to have norm $1$, then isn't it obvious that, say, $\sum \frac{e_n}{2^n}$ doesn't lie in the span of the $e_i$? This follows in the special case that the $e_i$ lie in a Hilbert space and are taken to be orthonormal, but in general you can't conclude this. Why? The naive continuation of the naive argument is that since $B = \text{span}(e_1, e_2, ...)$ there is a linear functional $$e_j^{\ast} : B \to \mathbb{C}$$

which, given a finite sum $\sum c_i e_i$, takes the value $c_j$, so we ought to have $$e_j^{\ast} \left( \sum \frac{e_n}{2^n} \right) = \frac{1}{2^j}.$$

Indeed there is such a linear functional, but it is not guaranteed to be continuous! So the last step fails. (When $H$ is a Hilbert space the inner product can be used to write down this functional and it is of course continuous in this case.)

  • $\begingroup$ In this particular case the functionals $e_j^{\ast}$ are continuous so the naive argument works and we can take, say, $\sum \frac{x^n}{n!}$ as an example of an element of the closure that doesn't lie in the space, but this general pitfall is worth looking out for. $\endgroup$ Jun 7, 2012 at 20:33

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