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For every linear functional $T$ on the space of convergent sequences in $\mathbb{R}$, how can I show it can be expressed $T(\{s_{n}\}) = \sum_{n \in \mathbb{N}} s_{n}T(e_{n})$ where $e_{n}$ are the canonical basis sequences?

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"lim" is such a linear functional, right? –  GEdgar Dec 3 '11 at 3:04
    
yes it is a linear functional –  Red Rover Dec 3 '11 at 3:24
    
possible duplicate of Properties of dual spaces of sequence spaces –  t.b. Dec 3 '11 at 3:39

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up vote 2 down vote accepted

You can see the claim in question in two steps, if you replace the space of "convergent sequences" with the space of "sequences convergent to zero."

Step 1: the claim is true for finitely supported sequences. This is easy from linearity of $T$ and since any finitely supported sequence is a sum (uniquely) of sequences of the form $e_n$.

Step 2: The finitely supported sequences are dense in the space you are considering. This follows because any sequence $\{c_n\}$ that converges to zero is the limit (with respect to the sup norm) of its finitely supported subsequences.

As GEdgar points out in the comments, the claim is false for the space of all convergent sequences. Since, in fact, this space splits as $c_0 \oplus \mathbb{C}$ (where $\mathbb{C}$ is identified with the space of constant sequences), it follows from this analysis that the linear functionals on the space of all convergent sequences are precisely the linear combinations of the ones you describe plus the one that sends a convergent sequence to its limit.

Note that without some condition of convergence, you can get a whole lot of functionals which cannot be explicitly described. For instance, $\ell^\infty$ has a very large dual, which includes things like Banach limits which have no simple description. (In fact, $\ell^\infty/c_0$ is an example of a Banach space on which there are lots of nonzero linear functionals, by the Hahn-Banach theorem, but which you can't really write down one without the axiom of choice!)

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The splitting $c = c_0 \,\mathrel{\oplus_\infty} \mathbb{C}$ you describe here is best understood when equipped with the maximum norm on the direct sum. The dual space then identifies with $\ell^1 \, \mathrel{\oplus_1} \mathbb{C} \cong \ell^1$. See also my answer in the possible duplicate thread I linked to. –  t.b. Dec 3 '11 at 3:55
    
@t.b.: Thanks for pointing to that answer out. –  Akhil Mathew Dec 3 '11 at 14:34

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