# 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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It's the continuous dual. – Alex Youcis Apr 14 '12 at 19:43
To expand a bit on Alex's comment, $X'$ is the space of bounded linear maps $f: X \to \mathbb{C}$ with norm $\| f \| := \sup_{\| x \| \leq 1} |f(x)|$. – Martin Wanvik Apr 14 '12 at 20:03

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.

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I object! When someone takes a course, and gets a problem to do, it is to help him learn something. When we provide a complete solution (in less than an hour!) it considerably reduces that learning. We should restrain ourselves, and provide merely a few hints (such as the comments did) and then wait for him to respond before going further. Norbert, you are not the first (nor the worst) offender in this regard, but I took this as a place to comment... – GEdgar Apr 14 '12 at 21:39

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