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Let $S \subset \ell^2$ be a infinite linearly independent compact subset of $\ell^2$ over $\Bbb C$

Let $V=span(S)$ and let $T:\overline{V} \to \Bbb C$ be a linear functional of $\overline{V}$

I would like to know if exist $v \in V$ such that $\lVert T\rVert = |T(v)|$

Thanks for any suggestion

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    And can you give an example of such set $S$, please? Because every infinite lineary independent subset of $l^2$ is unbounded, thus not compact. – 2017-02-23
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    @MichaelFreimann $S=\{\frac{1}{2^n}e_n : n \in \Bbb N\}$ and all lines between $\frac{1}{2^n}e_n$ and $\frac{1}{2^{n+1}}e_{n+1}$ – 2017-02-23
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    There exists a $v\in \overline{V}$ with this property. But not necessarily a $v\in V$. – 2017-02-23
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    @treedetective thanks for your comment – 2017-02-23
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    If you want more details tell me. Basically you just need to use James characterization of reflexivity via norm attaining functionals. – 2017-02-24
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    Your example $S$ is not compact since it doesn't contain $0$ but contains a sequence converging to $0$. If you take the lines between $2^{-n}e_n$ and $2{-n-1}e_{n+1}$ then the set is also not linearly independent. An example of an infinite compact linearly independent set would be $\{e_1+\frac1n e_n\}\cup \{e_1\}$. Also I'm sure you want a condition on $v$ beyond $v\in V$, for example $\|v\|=1$. Otherwise it is trivial that there exists a $v$ so that $|T(v)|=\|T\|$. In the case you want $\|v\|=1$ it is not always possible. – 2017-02-24

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