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I still didn't understand what is Second Dual Space. From class, I was told that: $X^{**}=\{g_{x}:x \in X\}$ which $g_{x}$ is a map from algebraic dual space $X^{*}$ to scalar field $K$ of vector space $X$. $g_x$ is defined by $g_x(f)=f(x)$. Then, I met $C$ which is called canonical mapping and map from vector space $X$ to second algebraic dual space $X^{**}$ defined by $Cx=g_x$. The book state that $C$ is injective but not always surjective. What I didn't understand is why $C$ isn't always surjective? From the definition of element in $X^{**}$ and from the statement that $C$ is injective, then $C$ must be surjective. Or I might be wrong define elements in $X^{**}$? And I'm using the book from Erwin Kreyszig - Introductory Functional Analysis with Applications

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  • $\begingroup$ $X^{**}=\{g_{x}:x \in X\}$ is not the definition of the second dual space. The second dual space is the dual of the dual space. $\endgroup$ – Aweygan Oct 25 '16 at 16:13
  • $\begingroup$ @Aweygan right, I'm sorry. This book said it's Second Algebraic Dual Space. Is it same with Second Dual Space? $\endgroup$ – user379677 Oct 25 '16 at 16:16
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You're wrong about the definition of $X^{**}$. From Kreyszig, page 107:

We may go one step further and consider the algebraic dual $(X^*)^*$ of $X^*$, whose elements are the linear functionals defined on $X^*$... and call it the second algebraic dual space of $X$.

He then goes on to explain that for any $x$, $g_x$ is an element of $X^{**}$.

Notably, however, there is nothing about the definition of $X^{**}$ which guarantees that every element $g \in X^{**}$, that is every $g:X^* \to \Bbb C$, has the form $g_x$ for some $x \in X$. In fact, if $X$ is an infinite dimensional normed space (and if we take the axiom of choice), then this is necessarily not the case.


For example: take $X\subset \ell^1$ to be the set of sequences for which all but finitely many elements are $0$.

Let $\{e_i\}_{i \in \Bbb N}$ be the canonical basis of $X$ (so for example, $e_1 = (1,0,0\dots)$); note that this is a Hamel basis for our space. Every functional $f$ is determined by its values over these elements, since $$ f(\sum_{i} \alpha_i e_i) = \sum_i \alpha_i f(e_i) $$ Thus, for every sequence $(\beta_i)_{i \in \Bbb N}$, there is an associated $f$ with $f(e_i) = \beta_i$.

Now, consider the subspace of $U \subset X^*$ consiting of $f \in X^*$ such that $$ \sup_{i \in \Bbb N}|f(e_i)| < \infty $$ We may define the linear map $g:U \to \Bbb C$ by $$ g(f) = \sup_{i \in \Bbb N} f(e_i) $$ and by the axiom of choice, we may write $X^* = U \oplus V$ for some subspace $V$, and we may extend $g$ to all of $X^*$ by defining $g|_V = 0$.

Note that $g \in X^{**}$, but there is no $x \in X$ such that for every $f\in X$, $g(f) = f(x)$.

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  • $\begingroup$ Do you happen to know whether the last assertion requires (part of) the axiom of choice? $\endgroup$ – Daniel Fischer Oct 25 '16 at 16:30
  • $\begingroup$ @DanielFischer It almost certainly does; I'll say so. $\endgroup$ – Omnomnomnom Oct 25 '16 at 16:31
  • $\begingroup$ So, It's not necessary that every $g \in X^{**}$ has the form $g_x$? If it's so, then how we define $g(f)$ if it has no form $g_x$? I'm really confused. $\endgroup$ – user379677 Oct 25 '16 at 16:39
  • $\begingroup$ It's hard to come up with an explicit example, but I added one in my latest edit. $\endgroup$ – Omnomnomnom Oct 25 '16 at 16:54
  • $\begingroup$ Ah, now I understand. So, element in $X^{**}$ is a function that maps from $X^*$ to $K$ and the definition is not always $g(f)=f(x)$. Thank you very much! $\endgroup$ – user379677 Oct 25 '16 at 17:08
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If we take $c_0 , \ell_1 , \ell_{\infty}. $ Then $c_0^{\ast} =\ell_1 , c_0^{\ast\ast}=\ell_1^{\ast}=\ell_{\infty}. $ Then the functional $\xi (x) =\sum_j x_j $ is not in the image $\kappa (c_0 )$ where $\kappa $ is the canonical mapping but $\xi \in c_0^{\ast\ast}=\ell_1^{\ast}=\ell_{\infty}.$

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  • $\begingroup$ One question, what's the dimention of $c_0, l_1, l_{\infty}$? Are they infinite dimentional vector spaces? $\endgroup$ – user379677 Oct 25 '16 at 16:40
  • $\begingroup$ $\mbox{dim} (c_0 ) =\mathfrak{c}$ $\endgroup$ – MotylaNogaTomkaMazura Oct 25 '16 at 16:49
  • $\begingroup$ In this answer, you're using continuous dual as opposed to the algebraic dual. $\endgroup$ – Omnomnomnom Oct 25 '16 at 16:54

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