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Assume we have $V = \mathbb{R}^{\mathbb{N}}$ over $\mathbb{F} = \mathbb{R}$.

If we look at the dual space to $\mathbb{R}^{\mathbb{N}}$, that is

$$A = \operatorname{Hom}_{\mathbb{R}}(\mathbb{R}^{\mathbb{N}},\mathbb{R})$$

what can we say about what elements this set will contain?

For example, if I interpreted my teacher correctly, he said that $$f = \sum_{n = 1}^{\infty} \Big(\frac{1}{2}\Big)^n e^*_n$$ is an element of $A$. But what does $f$ do with for example the element $$(1,2,4,\ldots)?$$

Just to clarify, here I denote $e_i$ as the unit vector $$(0,0\ldots,\underbrace{1}_{\text{position i}},0,0,\ldots) \in \mathbb{R}^{\mathbb{N}}$$

and $$e^*_i$$ as it's dual under the injection $$\varphi:V \to V^*$$ explicitly defined by $$V \ni v \longmapsto v^* \in V^*.$$

Edit: My teacher might have thought I meant the direct sum, not the direct product.

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    $\begingroup$ Are you looking at the algebraic or the topological dual space? $\endgroup$ Oct 5, 2023 at 13:47
  • $\begingroup$ 1) If he said that, he was wrong. 2) You didn't (and cannot) define $v^*.$ I think $e_i^*\in V^*$ is defined directly by $e_i^*(x)=x_i$, without reference to $e_i.$ (Btw, why "unit" vector? unit for which norm on $\Bbb R^{\Bbb N}$?) $\endgroup$ Oct 5, 2023 at 13:50
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    $\begingroup$ Just a rephrasing of what @AnneBauval said: the map $\varphi$ is not well-defined (actually you did not define it, but you can check that the obvious map is not well-defined). $\endgroup$ Oct 5, 2023 at 13:57
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    $\begingroup$ Check this post math.stackexchange.com/questions/1448024/… . The dual of $\mathbb R^{\mathbb N}$ is a huge space, and cannot be easily characterized. $\endgroup$ Oct 5, 2023 at 14:05
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    $\begingroup$ What do you call "defined the same way"? $\endgroup$ Oct 5, 2023 at 14:29

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The algebraic dual of $\mathbb R^{\mathbb N}$ is "huge" and hard to characterize, essentially for a similar reason why the (topological) dual of $\ell^\infty(\mathbb N)$ is huge: the possible behaviour of functions "at infinity" is so rich and arbitrary that you can take any functional defined on a subspace of sequences with a given behaviour at infinity (like converging sequences, periodic sequences, "quasi periodic" sequences,...) and extend it by the axiom of choice (or by Hahn Banach theorem in the case of the topological dual of $\ell^\infty$). There is just too much freedom on the behaviour of sequences at infinity (check this post for an concrete example of what I am talking about).

However (just an interesting remark), the topological dual of $\mathbb R^{\mathbb N}$ is much more well-behaved: it is simply the vector space $\operatorname{span}\{e_1^*,e_2^*,\dots\}$. The reason is that the topology on $\mathbb R^{\mathbb N}$ (thought as a topological vector space with topology induced by the linear functions $e_i^*$) is much, much weaker than that of $\ell^\infty$, and in particular the subspace of all sequences with only a finite number of nonzero elements is dense in $\mathbb R^{\mathbb N}$, so any linear continuous $\phi:V\to\mathbb R$ is uniquely determined by the values $\phi(e_i)$. So the topological dual space of $\mathbb R^{\mathbb N}$, unlike the algebraic dual space, does not see the behaviour at infinity of the sequences of $V$.


Edit. Replying to one of your comments to the question.

Let me explain more clearly what I meant in my comments. Let's say that instead of the $e_n^*$, you consider $\widetilde e_n^*$ in the following way:

  1. you take the family $\{e_n\}_{n\in \mathbb N}$
  2. you extend it to a basis $V'$ of $V$
  3. you impose that $\widetilde e_n^*(e_n)=1$ and $\widetilde e_n^*(v)=0$ for all $e_i\neq v\in V'$
  4. you extend $\widetilde e_n^*$ by linearity to all $v\in V$.

Then, my first point is that in general $e_n^*\neq \widetilde e_n^*$. So, even if it could look like that $\varphi$ is canonically definable, this is not the case. The functions $e_n^*$ are very "special" elements of the algebraic dual that can be written explicitly without needing to take a full basis $V'$, and the definition of any injective linear map $\varphi:V\to V^*$ does not follow canonically from the identification between $e_n$ and $e_n^*$: the latter is canonical, while the definition of the map $\phi$ necessarily depends on the choice of a basis of $V$, and there is no canonical way of assigning an element $v^*\in V^*$ to a given element $v\in V$ (unless $v$ is coincides with $e_n$ for some $n$, or it is a finite linear combination of the $e_n$'s.)

As a consequence, the map $v\mapsto\varphi(v)$ is "exotic" in the sense that you can not write it explicitly in terms of the entries of the sequence $v=(v_1,v_2,\dots)$, and it depends in a highly non explicit way on how you choose the basis $V'$, basically because any basis $V'$ itself is terribly non-explicit: all constructions of a (Hamel) basis $V'$ necessarily depend on the axiom of choice. (Maybe "exotic" is not the right word, but I hope I explained what I meant with it).

Finally, if you define $f$ replacing the $e_n^*$ by the $\widetilde e_n^*$, then $f$ is a well-defined element of $V^*$, because for any $v\in V$, there is only a finite number of $n$ such that $\widetilde e_n^*(v)\neq 0$, hence the infinite sum in the definition of $f$ reduces to a sum on a finite number of elements. But this new $f$ would be non explicit and dependent on the basis $V'$, as the definition of the $\widetilde e_n^*$ depends on the whole basis $V'$.

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$\def\NN{\mathbb N}\def\RR{\mathbb R}$ The set $\RR^\NN$ is the set of all real sequences. As such the $f$ your prof gave you (which very likely had a $2^n$ in the denominator) is not well-defined, as you correctly note.

Such $f$ makes if instead the domain is $\ell^\infty(\NN)$, the space of bounded sequences. This is the kind of things one studies in Functional Analysis.

As for the dual of $\RR^\NN$, we are talking about the algebraic dual here. I'll be happy to be proven wrong, but I don't think there is a nice characterization of this dual. Note that $R^N$ is a vector space of uncountable dimension, so any basis is uncountable and cannot be expressed in a concrete way; and the same is true of its dual. In fact, the dual has dimension strictly greater than that of $\RR^\NN$.

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    $\begingroup$ The map $f$ would be defined on $\ell^{\infty}$ if the $\frac{1}{2}$ factor were replaced with $\frac{1}{2^n}$. This must be a typo in the original question. $\endgroup$
    – Didier
    Oct 5, 2023 at 14:30
  • $\begingroup$ Indeed, thanks for noticing. I'll edit that in. $\endgroup$ Oct 5, 2023 at 14:31
  • $\begingroup$ Yes, you are ofc right, I´ve edited the question. (Regarding $(\frac{1}{2})^n$ and not $(\frac{1}{2})$). $\endgroup$
    – Ben123
    Oct 5, 2023 at 14:57

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