Let $\overline{X}$ be Banach space and $X$ be dense subset of it. Show that dual space of X and dual space of $\overline{X}$ are isomorphic.
Why these are isomorphic?
I don't know how to prove it.
[Math] The dual space of normed vector space $X$ is isomorphic to the dual of its completion
banach-spacesduality-theoremsfunctional-analysisnormed-spaces
Related Solutions
For future students, here is a more general result:
Let $X$ and $Y$ be normed linear spaces, and let $B(X,Y)$ denote the collection of all bounded linear operators from $X$ to $Y$ endowed with the operator norm. Show that $B(X,Y)$ is a normed linear space, and $B(X,Y)$ is a Banach space whenever $Y$ is a Banach space. The vector operations in $B(X,Y)$ are defined pointwise, i.e. $(A+B)(x)=Ax+Bx$, and $(\alpha A)(x)=\alpha (Ax)$. (Notice that in your case $X'=B(X,\mathbb{C})$ and $\mathbb{C}$ is a Banach space)
It is clear that linear operators form a linear space. To show that $B(X,Y)$ is a linear subspace, it is enough to show the closure to addition and scalar multiplication. But these follow easily from the properties of a norm (the fact that the operator norm satisfies all the properties of a norm for bounded functionals is an easy exercise that follows from properties of supremums in $[0, \infty)$) , namely for any $A,B \in B(X,Y)$ and $\lambda \in \mathbb{C}$ $$\|A+B\| \leq \|A\|+\|B\| < \infty$$ $$\|\lambda A\|=|\lambda| \cdot \|A\| < \infty$$ Thus, $B(X,Y)$ is a normed linear space.
Now assume that $Y$ is a Banach space. Let $\{A_i\}$ be a Cauchy sequence in $B(X,Y)$, i.e. $\forall \, \epsilon >0$, $\exists \, N \in \mathbb{N}$ such that $\forall \, m,n > N$, $\|A_n-A_m\|< \epsilon $. Let $x \in X$ be arbitrary. Let $\epsilon>0$ be arbitrary. If $x=0$, then $$\|A_nx-A_mx\|=0<\epsilon.$$ If $x \neq 0$, choose $N$ such that $\|A_n-A_m\|< \frac{\epsilon}{\|x\|}$. Then by a property of the operator norm, $\forall \, m,n > N$, \begin{equation} \begin{split} \|A_nx-A_mx\| & = \|(A_n-A_m)x\|\\ & \leq \|(A_n-A_m)\| \cdot \|x\|\\ & < \frac{\epsilon}{\|x\|} \cdot \|x\|\\ & = \epsilon\\ \end{split} \end{equation}
Thus, in both cases $\{A_nx\}$ is a Cauchy sequence in $Y$. Since $Y$ is a Banach space, it is convergent to some element in $Y$. Call that element $Ax$, i.e. $$\lim_{n \rightarrow \infty} A_nx=Ax$$ Since $x$ was arbitrary, $Ax$ is defined for any $x \in X$. Thus, $A$ is a map from $X$ to $Y$ defined by $x \rightarrow Ax$. We need to show that $A$ is linear, bounded, and $A_n \xrightarrow{n \rightarrow \infty} A$ in the operator norm. Notice that $A$ is linear, since by linearity of $A_n$ we get that for any $x_1, x_2 \in X$, $\lambda \in \mathbb{C}$, \begin{equation} \begin{split} A(x_1+x_2) & = \lim_{n \rightarrow \infty} A_n(x_1+x_2)\\ & = \lim_{n \rightarrow \infty} (A_nx_1+A_nx_2)\\ & = \lim_{n \rightarrow \infty} A_nx_1+\lim_{n \rightarrow \infty} A_nx_2\\ & = Ax_1+Ax_2\\ \end{split} \end{equation} \begin{equation} \begin{split} A(\lambda x_1) & = \lim_{n \rightarrow \infty} A_n(\lambda x_1)\\ & = \lim_{n \rightarrow \infty} \lambda \cdot A_nx_1\\ & = \lambda \lim_{n \rightarrow \infty} A_nx_1\\ & = \lambda\cdot Ax_1\\ \end{split} \end{equation}
Now recall that Cauchy sequences are bounded. Thus, $\forall \, n$, $\|A_n\|<C$ for some $C \in \mathbb{R}$. Using this fact, we can see that $A$ is bounded, since by continuity of a norm: \begin{equation} \begin{split} \|A\| & =\sup_{\|x\| \leq 1} \|Ax\|\\ & =\sup_{\|x\| \leq 1} \|\lim_{n \rightarrow \infty} A_nx\|\\ & =\sup_{\|x\| \leq 1} \lim_{n \rightarrow \infty} \|A_nx\|\\ & =\sup_{\|x\| \leq 1} \limsup_{n \rightarrow \infty} \|A_nx\|\\ & \leq \sup_{\|x\| \leq 1} \limsup_{n \rightarrow \infty} \Big(\|A_n\|\cdot \|x\|\Big)\\ & \leq \sup_{\|x\| \leq 1} C \cdot \|x\|\\ & = C \sup_{\|x\| \leq 1} \|x\|\\ & \leq C \\ \end{split} \end{equation}
Finally, we want to show that $A_n \xrightarrow{n \rightarrow \infty} A$ in the operator norm. Let $\epsilon > 0$ be arbitrary. Recall that for an arbitrary $x \in X$, we have $$\|A_nx-A_mx\| \leq \|(A_n-A_m)\| \cdot \|x\|$$ Since $\{A_n\}$ is Cauchy, choose $N$ big enough such that for all $n,m \geq N$, $\|(A_n-A_m)\| < \epsilon$. Then the above inequality turns into $$\|A_nx-A_mx\| \leq \epsilon \cdot \|x\|$$ Now by continuity of a norm, we can take limit on both sides as $m$ goes to infinity to obtain $$\|A_nx-Ax\| \leq \epsilon \cdot \|x\|$$ Now taking supremum on both sides over all $x$ such that $\|x\| \leq 1$ yields $$\sup_{\|x\| \leq 1}\|A_nx-Ax\| \leq \epsilon$$ But this is equivalent to saying that for all $n \geq N$, $$\|A_n-A\| \leq \epsilon$$ And since $\epsilon$ was arbitrary, this implies that $$A_n \xrightarrow{n \rightarrow \infty} A$$ in the operator norm. Thus, we conclude that $B(X,Y)$ is a Banach space.
You can construct the completion of a normed vector space $X$ by using a construction of the completion of a general metric space and then defining a normed vector space structure on that completion.
But there's a better way, or at least a way that a lot of people might regard as simpler and more elegant: If $X$ is a normed vector space then the dual $X^*$ is a Banach space, and $X\subset X^{**}$ just as if $X$ were a Banach space; the completion of $X$ is just its closure in $X^{**}$.
(More precisely, the completion of $X$ is the closure of $j(X)$, where $j:X\to X^{**}$ is the canonical injection; so $j$ is exactly the $T$ you ask for.)
Best Answer
Apparently, $X$ is a dense subspace of $\bar X$, in order to define its dual.
Clearly, if $\ell\in X^*$, then $\ell$ extends uniquely, by a standard density argument to an $\bar\ell\in \bar X^*$, and clearly $\|\bar\ell\|=\|\ell\|$. Inversely, if $\bar\ell\in\bar X^*$, then its restriction to $X$ is a bounded linear functional on $X$.