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!

  • $\begingroup$ It's the continuous dual. $\endgroup$ Apr 14 '12 at 19:43
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    $\begingroup$ To expand a bit on Alex's comment, $X'$ is the space of bounded linear maps $f: X \to \mathbb{C}$ with norm $\| f \| := \sup_{\| x \| \leq 1} |f(x)|$. $\endgroup$ 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 $\sup\{\Vert f_n\Vert:n\in\mathbb{N}\}\leq C$. 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.

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    $\begingroup$ 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... $\endgroup$
    – GEdgar
    Apr 14 '12 at 21:39
  • $\begingroup$ Nice detailed proof, thanks! Small typo: I guess the last line should be $\|f_n-f\|$ instead of $\|f_n-f_m\|$. $\endgroup$
    – dietervdf
    Oct 9 '16 at 19:41
  • $\begingroup$ @dietervdf, thank you! Typo fixed. $\endgroup$
    – Norbert
    Oct 9 '16 at 19:50
  • $\begingroup$ Can you explain why we needed $\limsup$ when proving $f$ is bounded? $\endgroup$ Sep 16 '20 at 21:29
  • $\begingroup$ @ASlowLearner, because we don't know for sure that the limit exists $\endgroup$
    – Norbert
    Sep 16 '20 at 22:21

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


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.

  • $\begingroup$ Can you explain why we needed $\limsup$ when proving $A$ is bounded? $\endgroup$ Sep 16 '20 at 21:29
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    $\begingroup$ @ASlowLearner we don't yet know $\lim_{n\rightarrow \infty}\lVert A_n\rVert$ exists, so use $\lim\sup$ instead. $\endgroup$
    – Flashhh
    Jun 25 '21 at 5:10

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