# Family of linear functionals on separable Banach space coinciding with norm when evaluated in one element

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!

• It's a consequence of the Hahn-Banach theorem. See this, also. Commented May 11, 2019 at 7:04
• @DavidMitra Thanks, so this answers existence (I edited my question accordingly), what about an explicit construction? Is this possible? Commented May 11, 2019 at 7:31
• 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. Commented May 11, 2019 at 8:09

## 1 Answer

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.

• Thank you very much, that's what I was looking for! Commented May 13, 2019 at 10:16