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Let $X$ be a (real or complex) vector space, let $X^{*}$ denote the vector space of all linear functionals defined on $X$, and let $X^{**}$ denote the vector space of all linear functionals defined on $X^*$.

Let the map $C \colon X \to X^{**}$ be defined as $$C(x) \colon= g_x \ \ \ \mbox{ for all } x \in X,$$ where $g_x$ is a functional on $X^*$ defined as $$g_x(f) \colon= f(x) \ \ \ \mbox{ for all } f \in X^*.$$

Now my question is, (how) is the map $C$ injective, especially when $X$ is infinite-dimensional?

I can show that $g_x$ is linear, and so is $C$.

Saaqib Mahmood
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1 Answers1

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I. If $g_x=g_y,$ then $f(x-y)=0, \forall f\in X^*.$
II. Every vector-space has a basis. So take a basis $\beta$ for $V.$ For every $z_0\in\beta,$ define $f_{z_0}(x):=c_{z_0}$ where $x=\sum_{z\in\beta}c_z*z,$ then $f_{z_0}\in X^*.$
III. If $f_{z_0}(x)=0, \forall z_0\in\beta,$ then $x=0.$

If any error occurs, feel free to tell me; hope this helps.

Community
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awllower
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  • the ** II. ** part of your proof is not clear. – Saaqib Mahmood Mar 26 '15 at 06:59
  • Express $x$ as a linear combination of basis elements, and define $f_z(x)$ to be the coefficient of $z?$ – awllower Mar 26 '15 at 07:01
  • expressing $x$ as a linear combination of some (finite number) of basis elements is fine, but can you please elaborate on how you would like the map to be defined? – Saaqib Mahmood Mar 26 '15 at 07:15
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    If the question is talking about bounded functionals, this method does not work. In that case you need Hahn-Banach. – Vincent Boelens Mar 26 '15 at 21:04
  • @SaaqibMahmuud How about the edited version, in which the notation is somewhat modified? And thanks to Vincent Boelens for pointing out the need for Hahn Banach, in the case of bounded functionals. – awllower Mar 27 '15 at 00:26