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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.2012-08-06

2 Answers 2

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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 $a$ metri$c$ $b$ut not a norm?2012-08-06
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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.2013-05-16