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?

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    $\begingroup$ Don't you mean "with the metric induced by the (semi)norms"? Completeness is a metric property, not a topological property (i.e. there exists a homeomorphism between a complete space and an incomplete space). $\endgroup$ – Nap D. Lover Apr 11 '16 at 19:04
  • $\begingroup$ Do you want to demand some kind of additional properties on this norm? Otherwise you could just take a linear isomorphism between $C^\infty(\mathbb{T})$ and your favorite separable infinite-dimensional Banach space and use the induced norm... $\endgroup$ – Eric Wofsey Apr 11 '16 at 19:14
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    $\begingroup$ For instance, the new norm might induce a completely different topology than the original metric. Operations like multiplication, differentiation, integration, might not be continuous under the new norm. Is that really what you want? $\endgroup$ – Nate Eldredge Apr 11 '16 at 19:31
  • $\begingroup$ At first I was thinking if $C^p(T)$ for $p<\infty$ with its usual metric could be isomorphic to $C^\infty(T)$ with another metric. So I think that atleast these operations should be continuous. $\endgroup$ – Chris Apostol Apr 11 '16 at 19:38
  • $\begingroup$ I added a precision in your question, hope this does not alter your intent. $\endgroup$ – Duchamp Gérard H. E. Jul 22 '17 at 9:22

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.

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