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So I was wondering what the dimension of the $C^2$ as a $\mathbb{C}$ vector space was. I could kinda figure out that the $C^\omega$ probably had a countable infinite dimension, while the space of all functions was uncountable.

This question came up because in physics it seems very common to explicitly index a basis of the $C^2$ and even sum over it, but I haven't seen a proof so far that there is a countable set of functions that create the $C^2$ through linear operations.

I tried a few things like looking at the derivative of all vectors of a basis of it, but I didn't manage to prove that those create the $C^1$, similarly I didn't manage to find an injection to the analytic functions.

Sorry if my formal language is off, I learned both analysis and algebra in German :/

nmasanta
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Chalky
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    Can you say precisely what you mean by $C^2$? – Eric Wofsey Jun 28 '19 at 02:23
  • Do you mean, twice continuously differentiable? – Gerry Myerson Jun 28 '19 at 03:04
  • Maybe you know this already, but if you're talking about a function space, note that "basis" often refers to a set of vectors whose linear span in dense in the space (e.g., the maps $\chi_n(x) = e^{2\pi i n x}$ for $n\in \mathbb{Z}$, in the space of "reasonable" functions with Fourier series expansion), whereas "dimension" refers to the cardinality of a set of linearly independent vectors whose linear span is the space itself. – anomaly Jun 28 '19 at 03:22
  • If you are using $C^2$ to mean twice continuously differentiable, then you should be able to work out an answer from https://math.stackexchange.com/questions/664084/the-dimension-of-the-real-continuous-functions-as-a-vector-space-over-mathbbr – Gerry Myerson Jun 28 '19 at 03:28

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