# Banach Reflexive Spaces

Let be $E$ a vector space and let be $$J:E\rightarrow E^{**}$$ $$x\mapsto J(x)\in E^{**}$$ a application that stat a relation between $E$ and $E^{**}$. My lecture say that expression $\langle J(x), f\rangle = f(x)$ for all $f\in E^*$ define a $J(x)$.

My question is: Why not exist any function $f$ such that $\langle J(x), f\rangle \neq f(x)$?

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As it seems to me that more than @Norbert's few words are needed, I will try to expand his answer. In my eyes, the first thing we need to understand is $E^{**}$. By definition, it is $(E^*)^*$, right? So an element $f \in E^{**}$ is a continouos, linear map $f \colon E^*\to \mathbb K$ ($\mathbb K$ denoting the ground field). That is, $f$ maps elements of $E^*$ (recall, these are maps $E \to \mathbb K$) to scalars. If we try to imagine such maps, the most basic of them are the evaluation maps $\def\ev{\mathrm{ev}}\def\norm#1{\left\|#1\right\|}\ev_x$, which do nothing but evalutating a given $\phi \colon E \to \mathbb K$ at a point $x \in E$. That is $\ev_x(\phi) := \phi(x), \qquad x \in E, \phi \in E^*$ That is, as one can see by noting $\left|\ev_x(\phi)\right| = \left|\phi(x)\right| \le \norm x \norm \phi$ a continuous, linear map $\ev_x \colon E^* \to \mathbb K$, so an element $\ev_x \in E^{**}$. For each $x\in E$, we have given $\ev_x \in E^{**}$, so we have a map $J\colon x \mapsto \ev_x$, so your $J$ is not any relation $E \to E^{**}$, but the map which associates with each $x \in E$ the functional "evalution at $x$" on $E^*$.
If $F$ is any topological vector space, $x \in F$ and $\phi \colon F \to \mathbb K$ linear and continuous, it is very common to denote the value $\phi(x)$ by $\langle \phi, x\rangle$ and the map $\langle\cdot,\cdot\rangle\colon F^* \times F \to \mathbb K$ is called the natural pairing of $F$ and $F^*$.
Applying this notation to $F = E^*$, we have for $x\in E$, $f\in E^*$: $\langle \ev_x, f\rangle = \ev_x(f) = f(x)$ and using the definition of $J$, we have for $x \in E$ that $J(x) = \ev_x$ and hence $\langle J(x), f\rangle = \langle \ev_x, f\rangle = f(x)$ so by the very definition of $J$ and $\langle\cdot,\cdot\rangle$, this holds for all $f\in E^*$, so there cannot be any $f$ violating this.
Denote $F=J(x)$, and $H=E^*$, then $F\in H^*$. So $F$ is a linear functional on $H$. Hence to define $F$ it is enough to know all the values $F(f)$ for all $f\in H$. This values are equal to $$F(f)=J(x)(f)\overset{\text{this is just a notaion}}{=}\langle J(x), f\rangle\overset{\text{this is definition of J for a given x} }{:=}f(x)$$