I am sure that this is well known in the right places, but: Is the C\* completion of a star nuclear Fréchet algebra a nuclear C\* algebra? (Suppose that the C\* norm is continuous with respect to the Fréchet topology.)


[Basic, but commutative, example to explain topology - smooth functions on a compact manifold completed to continuous functions.]


To explain more: I expect that if the result was true, all that would be needed would be that the product and star were continuous. If we take continuous seminorms $\|.\|\_1 \le \|.\|\_2 \le \dots$ then likely we should have $\|a\,b\|\_n\le C\, \|a\|\_m\,\|b\|\_m$ where $m,C$ just depend on $n$. By C\* completion, this would be via some star representation of the algebra on a Hilbert space $H$, so that the map from the algebra to $B(H)$ was continuous. 



Is a C* completion of a nuclear Fréchet algebra a nuclear C* algebra?
Counterexample.


I think the $C^\star$ algebra of the Free group on two generators $F\_2$ is not a nuclear $C^\star$ algebra, same for the reduced group $C^\star$ algebra of $F\_2$.


The finite support functions $c\_f (F\_2)$ is a dense subagebra of $C^\star(F\_2)$, but is not Frechet.


Let $\ell$ be the word length of $F\_2$. Let $\omega=e^\ell$, the exponentiated word length function on $F\_2$. Then $\omega(gh) \leq \omega(g) \omega(h)$ for $g, h \in F\_2$, and $\omega(e) = 1$. 


Form rapidly vanishing complex-valued functions on $F\_2$ using this "weight" $\omega$. I'll denote them by ${\cal S}^e(F\_2)$, ${\cal S}$ for "Schwartz functions". There is a natural Fr'echet space topology on ${\cal S}^e(F\_2)$, given by norms $\| \varphi \|\_n = sup\_{g \in F\_2} w(g)^n |\varphi(g)|$. ${\cal S}^e(F\_2)$ is naturally a Fr'echet algebra for convolution multiplication, because $\omega$ has the submultiplicative property. Also ${\cal S}^e(F\_2)$ is nuclear, since the exponentiated word lenght function satisfies a summability condition $\sum\_{g \in F\_2} {1\over{e^{pl(g)}}} < \infty$ for some $p>0$. And ${\cal S}^e(F\_2)$ is dense in the $C^\star$-algebra since it contains $c\_f(F\_2)$.


