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Let $P$ denote the vector space of polynomials $p:\mathbb R\to \mathbb R$. Is there an example of a norm $\|\cdot\|$ on $P$ so that the completion of $P$ under this norm contains functions of exponential growth, but not functions that grow super-exponentially (at infinity)?

My initial idea was to pick a basis (say monomials based at the origin), then represent each polynomial $p(x)=\sum_{k=0}^na_kx^k$ and place the weighted norm $$\|p\|:=\sum\frac{|a_k|}{k!}.$$ The completion of $P$ will then indeed contain functions of exponential growth, but also functions like $x\mapsto\exp(x^2)$, which I want to rule out. Is there a nice way of massaging this norm to get the desired result? Many thanks!

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    You want the coefficients to be small to have at most exponential growth. Try $$\lVert p\rVert = \sum k!\,\lvert a_k\rvert.$$ – 2017-02-23
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    @Fischer Nice one. Am I right that this given only exponential growth with basis smaller than $e$? Now I am interested in the question if it is possible to enable arbitrary exponential growth but nothing faster. – 2017-02-27
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    @DanielFischer Thanks, your comment helped me settle it. The norm $\|f\|:=\sup_x|f(x)\exp(-|x|\log(1+|x|))|$ does the job I want -- a Banach space containing polynomials and exponentials, but not functions of the form $x\mapsto\exp(x^{1+\varepsilon})$. – 2017-02-28

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Daniel already gave you the nice answer to choose the norm

$$\Vert p\Vert=\sum k!|a_k|.$$

This will indeed give you exponential growth with basis smaller than $e$. For example, you will find that

$$\Vert b^x \Vert =\frac1{1-\ln(b)},$$

which diverges for $b\rightarrow e$. Here is an argument, that you will not find a norm so that the completion enables arbitrary exponential growth but nothing faster. Assume you found such a norm $\Vert\cdot\Vert$ and let $f_n\in\Omega(n^x)$ with $\Vert f_n\Vert<\infty$. So we can define

$$g(x)=\sum \frac 1{2^n\Vert f_n\Vert} f_n(x),$$

with $\Vert g\Vert \leq 2$. But $g$ growth faster than any exponential.