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I have been finding it not so easy to understand the topology in $^{\prime}$ ie dual of $X$ . The most obvious on is the topology defined by operator norm topology .

But i am not able to get good intuition about defining weak topology and weak star on $X^{\prime}$ .

I have serious problem understanding the map $i_x : X \to X^{\prime\prime}$ . What i don't clearly understand is that when we define the weak topology on $X$ , then also $x \in X$ gets mapped to $f(x)$ , $f \in X^{\prime}$ , and i see that $i_x$ also does the same thing ?

Its am terribly getting confused . I need a help . Thanks

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I'm not sure if this answers your question, but:

The dual of $X^*$ is $X^{**}$ and consists of the bounded linear linear functionals on $X^*$. Now, given a point $x$ in $X$, pointwise evaluation at $x$ by elements of $X^*$ defines an element of $X^{**}$. So, in a sense, you can view $x$ as an element of $X^{**}$. This is what $i_x$ does - $i_x$ is the element $x$ viewed as a linear functional on $X^*$.

In general, though, $X^{**}$ may contain other functionals. The weak topology on $X^*$ is induced by elements of $X^{**}$ while the weak* topology is induced by elements of $X$:

A basic nhood of $0$ in the weak topology of $X^*$ has the form $ \{ x\in X^* | |f_1(x)|<\epsilon, |f_2(x)|<\epsilon, \ldots, |f_n(x)<\epsilon\} $ for some $\epsilon>0$ and $f_1,\ldots, f_n$ in $X^{**}$

A basic nhood of $0$ in the weak$^*$ topology of $X^*$ has the form $ \{ x\in X^* | |f_1(x)|<\epsilon, |f_2(x)|<\epsilon, \ldots, |f_n(x)|<\epsilon\} $ for some $\epsilon>0$ and $f_1,\ldots f_n$ in $i(X)$

Note that the only difference between the two is that with the weak* topology, you are only using the functionals on $X^*$ defined by elements of $X$.


With regards to your third paragraph, when you speak of the weak topology on $X$, the map $i$ doesn't come into play.

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    @Theorem Um, that's just the way it is :) Spaces where $X=X^{**}$ (that is, where $i$ is onto) are called [reflexive](http://en.wikipedia.org/wiki/Reflexive_space). An example of a non-reflexive space is $c_0$. Its first dual is $\ell_1$ while its second dual is $\ell_\infty$.2012-12-08
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First we must distinguish clearly the notation between dual algebraic $X^\prime$ and topological dual space $X^*$ of a normed espace $X$.

Secondly we must distinguish between the bidual algebraic $X^{\prime\prime}$ and topological bidual $X^{**}$ the same normed space X.

$X^\prime$ is the dual algebraic of $X$, i.e. the set of linear functional of $X$ continuous and no continuous. Then $ X^\prime =\left\{ \ell: X\to\mathbb{R}\quad \left| \quad \begin{array}{rl} \forall \alpha,\beta\in\mathbb{R}& \forall v_1,v_2\in V\\ \ell( \alpha\cdot v_1+\beta\cdot v_2)&=\alpha \ell(v_1)+\beta \ell(v_2) \end{array} \right. \right\} $

$X^*$ is the topological dual of $X$, i.e. the collection of all linear functional continuous ( equivalent to limited ) : $ X^*=\left\{ \ell\in X^\prime \quad \left| \quad \sup_{v\in X}\frac{|\ell(v)|}{\|v\|}<\infty \right. \right\} $ Them \begin{equation} \begin{split} X^{**}&=\left\{ L\in X^{*\prime} \quad \left| \quad \sup_{\ell\in X^*}\frac{|L(\ell)|}{\|\ell\|}<\infty \right. \right\} \end{split} \end{equation} and

\begin{equation} \begin{split} \mathcal{J}(X)&=\left\{ L\in X^{**} \quad \left| \quad \exists x\in X\; |\; L(\ell)=\ell(x) \right. \right\} \end{split} \end{equation}

I believe that this diagram might help you see the thing in total.!enter image description here