# 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?

• 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). Apr 11, 2016 at 19:04
• 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... Apr 11, 2016 at 19:14
• 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? Apr 11, 2016 at 19:31
• 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. Apr 11, 2016 at 19:38
• I added a precision in your question, hope this does not alter your intent. Jul 22, 2017 at 9:22

## 1 Answer

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.