Can $C^\infty(\mathbb{T})$ become a Banach space? Let $T$ be the unit circle and $C^\infty(\mathbb{T})$ the set of functions defined on $\mathbb{T}$
which have derivatives of every order. I know that $C^\infty(\mathbb{T})$ with the metric induced by the seminorms 
$$\sup_{t\in\mathbb{R}}|f^{(l)}(e^{it})|,l\geq 0$$ is complete (but not a Banach space with the seminorms themselves i.e. its locally convex structure cannot be defined by one norm).
Is there any chance that we can define some kind of norm on $C^\infty(\mathbb{T})$ in order to become a Banach space? 
 A: No, it is impossible to define a topology of $C^\infty(\mathbb{T})$ with a single norm if you want to preserve derivation (w.r.t. any non zero vector of the tangent space) and the product as a continuous operators (otherwise why take $C^\infty(\mathbb{T})$ ?) because - exponential - Taylor formula is not true on all $C^\infty(\mathbb{R})$. 
So, for example, call $D$ the derivation of $C^\infty(\mathbb{T})$ such that 
$$
D(f)[e^{it}]=-ie^{-it}\frac{d}{dt}(t\to f(e^{it}))  
$$
(all other invariant derivations are proportional). 
Then, if $D$ were continuous (i.e. bounded within the Banach structure), one would have, for all $t,h\in \mathbb{R}$
$$
\sum_{n\geq 0}\frac{h^n}{n!}D^n(f)[e^{it}]=f(e^{i(t+h)})
$$
which is not true for all $f\in C^\infty(\mathbb{T})$ (take e.g. any $f$ with support $\not= \mathbb{T}$). 
See a discussion on Taylor's formula 
here). 
A shorter proof Apply $zD$ (Euler operator) above to the family $z^n$ (i.e. $e^{nit}$ through the parametrization) and see that it has an unbounded spectrum which is impossible for a bounded operator.
