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$\mathbb{R}^n$ is a n-dim Hilbert space with dot product as inner product. The topology induced by the inner product is what is used in real analysis.

I was wondering as what kind of topological vector space $\mathbb{C}^n$ is regarded, especially in complex analysis? Do you consider some inner product on it and therefore it may be a Hilbert space, or do you consider some norm on it and therefore it may be a Banach space?

Thanks and regards!

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    Hilbert space is by definition a Banach space. So it is both.2011-05-07
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    See [Fabian's answer given here](http://math.stackexchange.com/questions/25157/how-to-prove-c-1-x-infty-leq-x-leq-c-2-x-infty/25163#25163) and follow the link he gives. Also, [Qiaochu's answer here](http://math.stackexchange.com/questions/37528/topology-on-the-general-linear-group-of-a-topological-vector-space/37552#37552) applies *mutatis mutandis*.2011-05-07
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    @Asaf: I know that. I just wonder if $\mathbb{C}^n$ is studied as a Banach space only, or further as a Hilbert space, in complex analysis. How are its norm and inner product defined?2011-05-07
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    Since you're an avid reader of Wikipedia, I really wonder what is unclear [in this article on inner products](http://en.wikipedia.org/wiki/Inner_product_space)?2011-05-07
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    @Theo: Thanks! I am "flattered" for being recognized as "an avid reader of Wikipedia".2011-05-07
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    Well, aren't many questions of yours (e.g. [this one](http://math.stackexchange.com/questions/36710/different-versions-of-riesz-theorems)) inspired by your lecture of this great resource?2011-05-07
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    I don't think my questions are duplicate of those linked in others' comment. I not just asked about the norm on $\mathbb{C}^n$, but also inner product on it.2011-05-07
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    Tim, see also [Sylvester's law of inertia](http://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia).2011-05-07
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    @Theo: How is Sylvester's law of inertia related to this discussion?2011-05-07
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    Well, it tells you that there is essentially only one scalar product on $\mathbb{C}^n$. (I just see that the complex case isn't treated there, but it is rather straightforward).2011-05-07

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You will find the answer to your question for example in S. Krantz's review paper on several complex variables, which you can get from here. As he explains in the first pages, the standard inner product on $\mathbb{C}^n$ is $$\langle z,w \rangle = \sum_{k=1}^n z_k \bar{w_k},$$ and this gives the same topology as when identifying $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ in the usual way.