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Let $f$ be a non zero continuous linear functional on a Banach space $X$. i.e. $$f:X\rightarrow\mathbb{R}$$ is linear and bounded. Let $E$ be any non empty closed convex set of $X$ such that $$sup_{x\in E} |f(x)|$$ is attained. Then show that the supremum is attained at some extreme point of E.

Now if the assumption includes that the set $E$ is also compact then the supremum should be attaind (by the extreme value theorem) further by Krein- Milman theorem we can guarantee that it's attained at some extreme point.

But what about my question ? Is there any resource or book to find the proof ?

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    $\begingroup$ What is your question? When linear functionals attain maxima on convex subsets? $\endgroup$ – s.harp Nov 6 '17 at 9:56
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    $\begingroup$ This strikes me as false. What if $X = \mathbb R$ and $f$ is constant? $\endgroup$ – Mees de Vries Nov 6 '17 at 10:30
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    $\begingroup$ @MeesdeVries The constant function isn't a linear function. $\endgroup$ – Demophilus Nov 6 '17 at 13:43
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    $\begingroup$ Indeed, I missed the absolute value under the supremum. However, you could modify the example to $E = \{0\} \times \mathbb R$. Then, the supremum is $0$, but $E$ does not have any extreme points. $\endgroup$ – gerw Nov 6 '17 at 15:15
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    $\begingroup$ @Pozz $\lbrace 1 \rbrace \times R$ is a counterexample for the linear functional $(x, y) \mapsto x$ on $R^2$, that is not trivial $\endgroup$ – SiD Nov 6 '17 at 16:03
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The statement as given is not true. Take $$ E = [0,1] \times \mathbb R $$ and $$ f(x,y)=x. $$ Then the supremum is attained on the line $\{1\}\times \mathbb R$. However $E$ has not extreme point at all.

You have to exclude by assumption that the minimum of $f$ is attained on a line, as lines do not tend to have extreme points.

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