Prove that $X'$ is a Banach space I'm taking a new course on functional analysis and meet with the following problem.
If $X$ is a normed space (not necessarily complete), then prove that $X'$ is a Banach space.
Definition: When the induced metric space is complete,the normed space is called a Banach space. I don't have idea here,in particular I don't know what does $X'$ stands for?
Regards!
 A: We can show more: 

If $X$ is a normed space and $E$ a complete normed space, then the vector space $L(X,E)$ of continuous linear maps from $X$ to $E$, endowed with the norm $\lVert T\rVert_{L(X,E)}:=\sup_{x\in X,x\neq 0}\frac{\lVert Tx\rVert_E}{\lVert x\rVert_X}$, is a Banach space. 

Let $\{T_n\}\subset L(E,F)$ a Cauchy sequence. Then for each fixed $x$, the sequence $\{T_nx\}\subset E$ is a Cauchy sequence, which converges by completeness to some element of $E$ denoted $Tx$. The map $x\mapsto Tx$ is linear; we have to check that it is continuous and that $\lVert T_n-T\lVert\to 0$. 
We get $n_0$ such that if $n,m\geq n_0$ then for each $x$ $\lVert T_nx-T_mx\rVert_E\leq\lVert x\rVert_X$ and letting $m\to+\infty$ we obtain $\lVert T_nx-Tx\rVert_E\leq\lVert x\rVert_X$ so $\lVert Tx\rVert\leq \lVert x\rVert+ \lVert T_{n_0}\rVert\lVert x\rVert$ and $T$ is continuous. 
Fix $\varepsilon>0$. We can find $N$ such that if $n,m\geq N$ and $x\in E$ then $\lVert T_nx-T_mx\rVert_E\leq \varepsilon\lVert x\rVert_X$. Letting $m\to \infty$, we get for $n\geq N$ and $x\in X$ that $\lVert T_nx-Tx\rVert_E\leq \varepsilon\lVert x\rVert_X$, and taking the supremum over the $x\neq 0$ we get for $n\geq N$ that $\lVert T-T_n\rVert_{L(X,E)}\leq \varepsilon$. 
A: For future students, below you can find a more general case, basically the one presented by @Davide Giraudo, but maybe a little more detailed.
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.
A: By definition $X'$ is a space of bounded linear functionals on $X$. More preciesly
$$
X'=\{f\in\mathcal{L}(X,\mathbb{C}):\Vert f\Vert<+\infty\}
$$
where $\mathcal{L}(X,\mathbb{C})$ is a linear space of linear functions from $X$ to $\mathbb{C}$ and
$$
\Vert f\Vert:=\sup\{|f(x)|:x\in X\quad \Vert x\Vert\leq 1\}
$$
In order to prove that $X'$ is complete consider Cauchy sequence $\{f_n:n\in\mathbb{N}\}\subset X'$. Fix $\varepsilon>0$. Since $\{f_n:n\in\mathbb{N}\}$ is a Cauchy sequence there exist $N\in\mathbb{N}$ such that for all $n,m>N$ we have $\Vert f_n-f_m\Vert\leq\varepsilon$. Consider arbitrary $x\in X$, then
$$
|f_n(x)-f_m(x)|=|(f_n-f_m)(x)|\leq\Vert f_n-f_m\Vert\Vert x\Vert\leq\varepsilon \Vert x\Vert
$$
Thus we see that $\{|f_n(x)|:n\in\mathbb{N}\}\subset\mathbb{C}$ is a Cauchy sequence. Since $\mathbb{C}$ is complete, there exist unique $\lim\limits_{n\to\infty}f_n(x)$. Since $x\in X$ is arbitrary we can define function
$$
f(x):=\lim\limits_{n\to\infty}f_n(x)
$$
Our aim is to show that $f\in X'$ and $\lim\limits_{n\to\infty}f_n=f$.
Let $x_1,x_2\in X$, $\alpha_1,\alpha_2\in\mathbb{C}$ then
$$
f(\alpha_1 x_1+ \alpha_2 x_2)=
\lim\limits_{n\to\infty}f_n(\alpha_1 x_1+ \alpha_2 x_2)=
$$
$$
\lim\limits_{n\to\infty}(\alpha_1 f_n(x_1) + \alpha_2 f_n(x_2))=
\alpha_1 \lim\limits_{n\to\infty}f_n(x_1) + \alpha_2 \lim\limits_{n\to\infty}f_n(x_2))=
\alpha_1 f(x_1) + \alpha_2 f(x_2)
$$
So we conclude $f\in\mathcal{L}(X,\mathbb{C})$. Since $\{f_n:n\in\mathbb{N}\}$ is a Cauchy sequence it is bounded in $X'$, i.e. there exist $C>0$ such that $\sup\{\Vert f_n\Vert:n\in\mathbb{N}\}\leq C$. Hence, for all $x\in X$ we have
$$
|f(x)|=
|\lim\limits_{n\to\infty}f_n(x)|=
\lim\limits_{n\to\infty}|f_n(x)|\leq
\limsup\limits_{n\to\infty}\Vert f_n\Vert \Vert x\Vert\leq
\Vert x\Vert\sup\{\Vert f_n\Vert:n\in\mathbb{N}\}\leq
C\Vert x\Vert
$$
Now we see that $\Vert f\Vert\leq C$, but as we proved earlier $f\in\mathcal{L}(X,\mathbb{C})$, so $f\in X'$.
Finally recall that for given $\varepsilon>0$ and $x\in X$ there exist $N\in\mathbb{N}$ such that $n,m>N$ implies
$$
|f_n(x)-f_m(x)|\leq\varepsilon \Vert x\Vert.
$$
Then let's take here a limit when $m\to\infty$. We will get
$$
|f_n(x)-f(x)|\leq\varepsilon \Vert x\Vert.
$$
Since $x\in X$ is arbitrary we proved that for all $\varepsilon>0$ there exist $N\in\mathbb{N}$ such that $n>N$ implies
$$
\Vert f_n-f\Vert=
\sup\{|f_n(x)-f(x)|:x\in X,\quad \Vert x\Vert\leq 1\}\leq
\varepsilon.
$$
This means that $\lim\limits_{n\to\infty} f_n=f$. Since we showed that every Cauchy sequence in $X'$ have a limit, $X'$ is complete.
This proof can be easily generalized up to the following theorem: If $X$ is a normed space and $Y$ is a Banach space, then the linear space of all bounded linear functions from $X$ to $Y$ is complete.
