Suppose $X$ is a possibly non-separable Banach space and let $X^*$ be its dual. Also, let $(f_n)_{n=1}^\infty$ be a sequence of [EDIT: linearly independent] norm-one functionals in $X^*$. Does there exist an element $x\in X$ such that $f_n(x) \in (0,\infty)$ for all $n$?
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You mean functionals $f_n:X\to F$ s.t. $f_n(x)>0$? Just clarifying notation. – Alex Becker Feb 24 '13 at 22:59
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@Norbert I think that given a sequence $f_n$, (s)he wants to find such an $x$. – Alex Becker Feb 24 '13 at 23:02
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You would have to add some extra hypothesis so as to exclude $f_0=-f_1$ for instance... – Olivier Bégassat Feb 24 '13 at 23:16
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Not necessarily. Consider the case $X=c_0$, the Banach space of all real sequences $x=(x_n)_{n\in \Bbb N}$ that tend to $0$ as $n$ tends to infinity. We define continuous linear functionals $\mathrm{ev}_n$ on $X$ that send $x$ to its $n$-th component $x_n$. Then let us put $f_0=\mathrm{ev}_0$ and for all $n\geq 1$, $$f_n=C(~\mathrm{ev}_{n}-2\mathrm{ev}_{n-1})$$ where $C>0$ is defined such that the $f_n$ have unit length (it should be independent of $n$). If there were $x\in X$ such that all the $\mathrm{ev}_n(x)$ are positive, then $x$ would grow faster than $2^n$, contradicting the fact that by definition $x_n$ tends to $0$ as $n$ tends to infinity.
Olivier Bégassat
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