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Let $(E,\|\cdot \|)$ be a normed vector space. I have to prove that, for fixed $x\in E$, it holds that $$\|x\|:=\sup \{ \langle f,x\rangle_{E' \times E} : f \in B_{E'}\} = \max \{ \langle f,x\rangle_{E' \times E} : f \in B_{E'}\}$$ where $B_{E'}=\{ f \in E' : \|f\|_{E'}\le 1\}$ and $\|f\|_{E'}=\sup\{|f(x)| : x \in E \text{ and} \|x\|\le 1\}$.

My idea is to show that there exists $f \in B_{E'}$ such that sup is attained, but I don't know how to prove it. Suggests?

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    I feel confused because to define the dual unit ball I need the dual norm, so the norm of $E$ that is defined using functional in the dual unit ball – 2017-01-11
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    Just to get you thinking a little: $x$ can be identified with the linear functional $x'' : E' \to K,x''(f)=f(x)$, where $K$ is the base field. This is a continuous linear functional on $E'$. It would attain its maximum on a compact, but the unit ball in $E'$ is not compact in the strong topology unless the space is finite dimensional. Can you work around that fact? – 2017-01-11
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    In order to show that the supremum is attained,you have to prove the existence of a certain functional. Do you know of any results which show the existence of a functional? – 2017-01-11

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