Functional Analysis Bollobas Problem 3.14

[low flower]

Im trying to doing the exercise of bollobas, in this in particular i have a lot of questions, firts i separated the elementes x_0 from the other points in X/K (clausure) and i use hanh banach to found the functional, then i use the supreme and the f(x_n) > sup f(x_0) for x_0 in K ( I used tha exist x_n who converges to x_0 but x_n is not in K), my dudes are why f is not 0 (f != 0) and why the supreme is reached and then f(x) = sup f(x_0) for x_0 in K for a x element of the frontier of K (i think i hace used propierte of a convex in a evn of R to conclude the supreme is reached, but why f != 0 ?)

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[silk rapids]

I don't think this exercise works. You need to further assume that $0\notin\partial K$. If $0\in\partial K$ (as in the case of e.g. ${(x,y):(x-1)^{2}+y^{2}\leq1}$ which you can show is closed and convex in $\mathbb{R}^{2}$) then any linear functional $f\in X^{*}$ would have $f(0)=0$ so $\sup_{x\in K}f(x)=f(0)=0$ implies $f=0$.

[rose sable]

Pear Category Theorem

[silk rapids]

Otherwise, you can assume $f$ is not 0 because when you use HB you wind up assuming something like $||F||=||f||$ or $f(x_{0})=||x_{0}||$ or something of the sort.

[rose sable]

Pear Category Theorem

[empty moat]

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