Canonical embedding of $X$ in $X^{\prime\prime}$ and functionals of $X^{\prime}$ 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 
 A: 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.
A: 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.!
