Prove that $X'$ is a Banach space

Question

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!

Answer 1

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 $\sup\{\Vert f\Vert:n\in\mathbb{N}\}\leq C$. 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 $$ \Vert f_n-f_m\Vert= \sup\{|f_n(x)-f(x)|:x\in X,\quad \Vert x\Vert\leq 1\}\leq \varepsilon. $$ 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.

Answer 2

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