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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'm not sure what youre asking. you want to prove some space isn't reflexive based on what? – yoyo Dec 17 '11 at 16:10
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|| – user18217 Dec 17 '11 at 16:11
up vote 6 down vote accepted

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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correct me if I am wrong but φ(af+bf)=|a+b|$\cdot$||f|| <= |a|$\cdot$||f||+|b|$\cdot$||f||. So φ is not a linear functional in that case. – user18217 Dec 17 '11 at 16:25
I'm taking your specific $f$ that exists by hypthesis and I'm only asking that for that one $f$ we have $\varphi(f) = \|f\|$. For $\lambda f$ in the linear span of $f$ put $\bar\varphi(\lambda f) = \lambda \|f\|$. This is linear and of norm $1$ on the span of $f$. Now extend $\bar{\varphi}$ to $\varphi \in X^{\ast\ast}$. – t.b. Dec 17 '11 at 16:27
Ok, thanks. Can you please explain the last sentence ? (from "Since") What is the cononical inclusion and what is ev_x ? – user18217 Dec 17 '11 at 16:34
@user: I write $\operatorname{ev}_x(g) = g(x)$ for the map $X \to X^{\ast\ast}$ given by evaluation (the one that is used in the definition of reflexivity). We know that $\|\varphi\| = 1$ so, if it is of the form $\varphi = \mathrm{ev}_x$ for some $x$, we must have $\|x\| = 1$ because $\mathrm{ev}$ is isometric. But this gives the contradiction $\|f\| = |\varphi(f)| = |\mathrm{ev}_x(f)| = |f(x)| \lt \|f\| \|x\| = \|f\|$. – t.b. Dec 17 '11 at 16:40
@user18217: It is very important to keep in mind that reflexivity is about that specific morphism $\operatorname{ev}: X \to X^{\ast\ast}$ given by $\operatorname{ev}_{x}(f) = f(x)$ being onto. In fact, the James space $J$ is an example of a Banach space for which $\operatorname{ev}: J \to J^{\ast\ast}$ is not onto but for which there exists an isometric and surjective map $T : J \to J^{\ast\ast}$. – t.b. Dec 17 '11 at 18:12

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