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I have been giving the following corollary while studying functional analysis

Let $X$ be a normed vector space. Then the evaluation map $$ev : X \to X'' , x \mapsto (f \mapsto fx) $$ is an isometry.

I know this is a consquence of the Hahn-Banach Theorem

I have been asked to find an example of a Banach space $X$ whose evaluation map is not surjective.

But since the evaluation is a distance persevering function and is automatically injective how would one find an example of a space whose evaluation map is not a surjective? I can't seem to think of a Banach space where this is true.
Maybe my understanding of the double dual is too limited ?

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    $\begingroup$ It may help to remind you that spaces where the evaluation map is surjective are called "reflexive spaces". $\endgroup$ Commented Nov 6, 2014 at 10:15

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You may try $l_1$, whose dual is $l_{\infty}$ by simply assigning

$\begin{equation*} l_1 \longrightarrow l_{\infty}^* \\ y \longmapsto w_y : l_{\infty} \longrightarrow \mathbb{R}\\ \qquad \qquad \quad x \longmapsto \sum_n x_ny_n \end{equation*}$

Now consider the map

$\begin{equation*} \lambda : c \longrightarrow \mathbb{R}\\ \lambda(x) = \lim_n x_n \end{equation*}$

Use Hahn-Banach to prove that it admits an extension

$\begin{equation*} \widetilde{\lambda} : l_{\infty} \longrightarrow \mathbb{R} \end{equation*}$

but does not exist any $y \in l_1$ such that $\widetilde{\lambda} =w_y$; hence we have

$\begin{equation*} l_1 \hookrightarrow l_{\infty}^* \end{equation*}$

which is not surjective, that is, $l_1 \neq l_{1}^{**}$

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Hint: The dual of $\ell^1$ is $\ell^\infty$, but any non-principal ultrafilter on $\bf N$ gives an element of the dual of $\ell^\infty$ which is not in $\ell^1$.

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Take for example $X=L^1(\Omega)$. And consider the Dirac functional in $L^\infty$, defined as the functional $\delta_x:C^0(\bar{\Omega})→\mathbb{R}$ with $\delta_x(f)=f(x)$ and continued to $L^\infty$ by Hahn-Banachs theorem. This is not in the image of the canonical map from $X$ to the bidual $X''$.

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$
    – Argha
    Commented Nov 6, 2014 at 10:46
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    $\begingroup$ @Argha The answer is very terse and could benefit from a little bit more explanation, but it's an answer... $\endgroup$ Commented Nov 6, 2014 at 10:55
  • $\begingroup$ @NajibIdrissi: I think this is better suited for comment not for answer.. $\endgroup$
    – Argha
    Commented Nov 6, 2014 at 10:57
  • $\begingroup$ If by "Dirac functional" you refer to a point mass or the distribution it generates, that is neither in $L^1(\Omega)'$ not in $L^\infty(\Omega)'$. If you refer to something else, what to? $\endgroup$ Commented Nov 6, 2014 at 14:48
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    $\begingroup$ Okay, that works (assuming the $x_0$ is a typo and you meant $f(x)$). But you should say that in the answer. $\endgroup$ Commented Nov 6, 2014 at 14:56

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