Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\langle J(x), f\rangle \neq f(x)$?

share|improve this question

2 Answers 2

up vote 0 down vote accepted

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.

share|improve this answer

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 $$ F(f)=J(x)(f)\overset{\text{this is just a notaion}}{=}\langle J(x), f\rangle\overset{\text{this is definition of J for a given x} }{:=}f(x) $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.