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Is there a statement that $X^*$ must be "bigger" than X (considering that X is general banach space, not necessarily Hilbert anymore, it's trivial in hilbert because Riesz representation)? Is there a theorem about the nature of this?

"larger" here means that there is a surjective map from $X^*$ to X.

I guess there's no other restriction like continuity imposed on 'surjective map', as I only want to check $X^*$ cannot be much smaller than $X$, like how $\mathbb{Q}$ in $\mathbb{R}$ does.

The more precise expression may be:'Is cardinality of the continuous dual $X^∗$ at least as large as that of X ?'(Thanks to the suggestion of David Gao)

What's the motivation of this question?

I learned that X can be embedded into $X^{**}$, so I guess I can say something about the relationship between X and $X^{*}$

Why I use 'surjective'(some textbook writes 'onto', 'full map'),here is the example:

eg: dual of $L^1[0,1]$ is $L^\infty[0,1]$, with $L^\infty[0,1] \subseteq \ne L^1[0,1]$. But I also fail to say 'does not exist a surjective map', as we have examples like: even numbers are contained in intergers but still have surjective map.

Cabbagebislikethis
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  • You might want to be more specific about a "surjective map". Do you mean any surjective map between the spaces as sets? Do you want the map to be linear? Do you want the map to be continuous and linear? – Theo Bendit May 13 '24 at 13:52
  • See e.g. https://math.stackexchange.com/q/35779/620957, https://en.wikipedia.org/wiki/Dual_space#Infinite-dimensional_case, https://en.wikipedia.org/wiki/Reflexive_space. It also depends on whether you mean the algebraic or continuous/topological dual space. – Milten May 13 '24 at 13:55
  • To get you started with a brief answer though: In the case of the algebraic dual and infinite dimensions, the linear dimensionality of $X^*$ is strictly greater than that of $X$. – Milten May 13 '24 at 13:58
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    @Milten That's not exactly what I ask for >_<. I've learned something like reflexive space but my question is: is it true that X cannot be much bigger than $X^$, under the meaning of surjective(like no surjective from $\mathbb{Q}$ to $\mathbb{R}$). As we may talk about infinite dimension Banach space, there's NO COUNTABLE BASIS. Here dual $X^$ means bounded(or equally, continuous) linear map from X to F(here we can take F to be $\mathbb{R}$), maybe I should clarify it. – Cabbagebislikethis May 13 '24 at 14:12
  • @Cabbagebislikethis So your question is whether the cardinality of the continuous dual $X^\ast$ is at least as large as that of $X$? – David Gao May 13 '24 at 14:54
  • If that’s the question, I should first say that that is certainly not true for general locally convex Hausdorff TVS. I would assume it’s probably true for Banach spaces, but it’s not clear to me how to prove this at the moment. – David Gao May 13 '24 at 15:20
  • (In case of $L^\infty[0, 1]$ and $L^1[0, 1]$, this is certainly true though. Both have cardinality continuum.) – David Gao May 13 '24 at 15:22
  • @DavidGao Yes, cardinality is what i'm looking for – Cabbagebislikethis May 13 '24 at 15:25
  • Generalizing the $L^\infty$ and $L^1$ example, for an infinite cardinal $\kappa$, it is clear that $\kappa \leq |L^\infty([0, 1]^\kappa)| \leq |L^1([0, 1]^\kappa)| \leq \kappa^{\aleph_0}$. So if $\kappa = \kappa^{\aleph_0}$, they are all equal. It is an interesting question to ask when $\kappa < \kappa^{\aleph_0}$, which of the three inequalities can be strict. – David Gao May 13 '24 at 16:21

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