Let $P(a,b)$ be the set of (finite degree) polynomial on $(a,b)$. As a consequence of Baire's theorem we know that that
Since $P(a,b)$ admits a countable Hamel's basis, then there is no norm that makes it a Banach space
Now, if I consider the set of analytic functions on $(a,b)$, does it admit a countable Hamel's basis? if the answer is no (as I expect), does it admit a complete norm?
The space $\mathcal A$ of analytic functions on $(a,b)$ has dimension $\mathfrak c$ (the continuum): note that $\exp(cx)$ for $c \in \mathbb C$ are linearly independent, while on the other hand the cardinality of the space is easily seen to be $\mathfrak c$ (e.g. each analytic function is determined by its Taylor series around a point of $(a,b)$). So as a vector space $\mathcal A$ is isomorphic to any other vector space with dimension $\mathfrak c$, including every separable Banach space, and (in principle) you can give it a complete norm from such a space. Of course the topology from this norm will not likely be anything useful in terms of analytic functions: you won't even be able to describe it explicitly.