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Let $\left(X,\|\cdot\|\right)$ be a Banach space. I need to show that if $\exists f:X \to K$ ($K$ is either the real or complex numbers) a bounded linear functional s.t $\forall x\in X \setminus \{ 0\} ,\, |f(x)|< \|f\| \cdot \| x \|$, so $X \neq X^{**}$ ($X^{*}$ is the dual space of $X$).

We got a hint: to use Hahn-Banach for a subspace of $X^{*}$. I thought about it and I don't have any good idea how to prove that, I would be glad to get some help.

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    I want to show that if the following condition exists the space is not reflexive: ∃f:X→K (K is either the real or complex numbers) a bounded linear functional s.t ∀x∈X∖{0},|f(x)|<||f||⋅||x||2011-12-17

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Hint: Let $f \in X^\ast$ be as in your hypothesis. Construct a linear functional $\varphi \in X^{\ast\ast}$ of norm $1$ such that $\varphi(f) = \|f\|$ by applying Hahn-Banach to a suitable functional defined on the linear span of $f$. Since the canonical inclusion is isometric $\varphi \neq \operatorname{ev}_{x}$ for all $x$ with $\|x\|=1$ by your condition, so $\mathrm{ev}: X \to X^{\ast\ast}$ is not onto.

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    I didn't know that, now it makes sense, thanks.2011-12-17