Proof of "Dual normed vector space is complete" http://en.wikipedia.org/wiki/Dual_norm
As in the introduction of dual norm by Wiki, it says dual normed space $X'$ is always
complete.
How to prove that? or at least explain that?
We all know the normed vector space is not always complete; if complete, all Cauchy sequences convergent. However, dual normed vector space is complete?
 A: Hint If $(f_n)$ is a Cauchy sequence of bounded functionals $f_n:V\to k$, take any $x$ and prove $f_n(x)$ is a Cauchy sequence in $k$. I'd take you assume your base field is complete, for example $k=\Bbb R$. 
A: 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.
