# When the linear operator attains its maximum on a convex set

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 ?

• What is your question? When linear functionals attain maxima on convex subsets? – s.harp Nov 6 '17 at 9:56
• This strikes me as false. What if $X = \mathbb R$ and $f$ is constant? – Mees de Vries Nov 6 '17 at 10:30
• @MeesdeVries The constant function isn't a linear function. – Demophilus Nov 6 '17 at 13:43
• 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. – gerw Nov 6 '17 at 15:15
• @Pozz $\lbrace 1 \rbrace \times R$ is a counterexample for the linear functional $(x, y) \mapsto x$ on $R^2$, that is not trivial – SiD Nov 6 '17 at 16:03

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.