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I am looking at a space of holomorphic functions defined on an unbounded set in C. This space contains unbounded functions (but bounded on compact subsets).

What is the classical/typical norm one would use on such a function space?

Thanks

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It may be worth noting that the space you describe cannot be given any complete, submultiplicative norm, since it contains functions $f$ such that $f(\Omega)$ is unbounded, while all elements of a Banach algebra must have compact spectrum. –  user16299 Aug 6 '12 at 7:15

2 Answers 2

you can use $$\partial(x,y)=\frac{d(x,y)}{1+d(x,y)}$$

Next set $$\partial_k(f,g)=\sup\limits_{z\in E_k}\partial(f(z),g(z))$$ which may be described as distance between $f$ and $g$ on some compact set $E_k$ and finally set $$\rho(f,g)=\sum_{k=1}^{\infty}2^{-k}\partial_k(f,g)$$

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Surely this is a metric but not a norm? –  user16299 Aug 6 '12 at 7:11

I don't know of any natural way to make the set of all holomorphic functions into a Banach space. You could pick a compact set $E$ and define $\|f\|=\sup_E |f|$: this is a norm, but the function space is not a complete with respect to it. Not of much use.

One option is to use a weighted Bergman space with a rapidly decaying weight, such as the Fock space. But no matter how quickly the weight decays, this space will not contain all holomorphic functions.

The other option is to use a family of norms instead of a single norm, namely $\|f\|_k=\sup_{E_k}|f|$ where $E_1\subset E_2\subset \dots$ are compact sets whose union is your open set. This is a locally convex topological vector space whose topology can be given by a translation-invariant complete metric (as in the answer by @Patience). Just the nicest thing you have, short of a Banach space.

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inte‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌resting. –  user59671 May 16 '13 at 20:32

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