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!
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$\begingroup$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.
$\endgroup$ 13 $\begingroup$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 $\lVert T-T_n\rVert_{L(X,E)}\leq \varepsilon$.
$\endgroup$ $\begingroup$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.
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