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let $B$ be a real, separable and infinite-dimensional Banach space and let $x_1$, $x_2$, $x_3$, $\dots$ be a dense subset of $B$. In the proof of Lemma 2.1 on p. 2 in this article, the author takes bounded linear functionals $F_n$ such that $\lVert F_n \rVert=1$ (inspecting the rest of the proof, this norm should be the operator norm) and $F_n (x_n) = \lVert x_n \rVert_B$.

My question is: Are there explicit ways to construct such a family of functionals, for example if $B$ is $C([0,1])$ or if $(x_n)$ is a Schauder basis of $B$?

Thanks a lot for your help!

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  • $\begingroup$ It's a consequence of the Hahn-Banach theorem. See this, also. $\endgroup$ Commented May 11, 2019 at 7:04
  • $\begingroup$ @DavidMitra Thanks, so this answers existence (I edited my question accordingly), what about an explicit construction? Is this possible? $\endgroup$
    – herrsimon
    Commented May 11, 2019 at 7:31
  • $\begingroup$ I'd expect in general to not be able to give nice explicit forms for these functionals. However a comment is that both types of space you mention are separable and in this case you don't need the axiom of choice to prove Hahn-Banach and so the proof of the result in that special case does in principle give you a description of the functionals involved. $\endgroup$ Commented May 11, 2019 at 8:09

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It is possible to construct such functional in case of hilbert spaces: just let $F_n(x_n)=\|x_n\|$ and $F_n(y)=0$, whenever $y\in\langle x_n\rangle^\perp$.

In case of $C[0,1]$ it is also possible. Let $a_n\in[0,1]$ be the point, such that $x_n(a_n)=\alpha_n\|x_n\|_\infty$, where $|\alpha_n|=1$. Then functional $F_n(x)=x(a_n)/\alpha_n$ is what you need.

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  • $\begingroup$ Thank you very much, that's what I was looking for! $\endgroup$
    – herrsimon
    Commented May 13, 2019 at 10:16

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