Is a C* completion of a nuclear Fréchet algebra a nuclear C* algebra? 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. 
 A: 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 $c_f(F_2)$.
