Let $A$ be a subset of $\mathbb{R}^n$, and denote by $C(A)$ the space of complex-valued continuous functions defined on $A$. We know that if $A$ is compact then we can define a norm on $C(A)$ so that it is a Banach space. If $A$ is open in $\mathbb{R}^n$ then there exists a topology on $C(A)$ that makes it a Frechet space but not normable (see, for example, page 33-34 of Rudin's Functional Analysis).


Now suppose $A$ is open in $\mathbb{R}^n$. My question is if there exists a topology on $C(A)$ so that the topological vector space $C(A)$ is normable? I don't have any preference for the answer, so either yes or no is good to me.


Similarly I would like to ask the same question for $C^k(A)$, where $k\in\mathbb{N}$, for $C^\infty(A)$, and for $C^\infty(M)$ where $M$ is a smooth manifold (you may put any assumption on $M$ as you like, compact, closed, or has a boundary, etc).


P.S. I know the [Kolmogorov's normability criterion](https://en.wikipedia.org/wiki/Kolmogorov%27s_normability_criterion), but I guess there exist some topologies on $C(A)$ other than the one above, assuming $A$ is open in $\mathbb{R}^n$. So I am not sure if there is some topology on $C(A)$ that would make it normable.



About the normability of the space of continuous functions
Every real or complex vector space can be equipped with a norm (at least under the axiom of choice): Take a Hamel basis $B$ with coefficient functionals $\varphi\_b$ and define $\|x\|=\sum\limits\_{b\in B} |\varphi\_b(x)|$.


This norm however, is hardly ever of any use. For all your examples, natural additional assumptions would be completeness of the norm and the continuity of all evaluations $\delta\_x: f\mapsto f(x)$. The closed graph theorem then implies that the norm topology would be equal to the natural Frechet topology -- which cannot be since in the examples you have proper (non Banach) Frechet spaces.


