As in the introduction of dual norm by Wiki, it says dual normed space $X'$ is always complete.

How to prove that? or at least explain that?

We all know the normed vector space is not always complete; if complete, all Cauchy sequences convergent. However, dual normed vector space is complete?

up vote 3 down vote accepted

Hint If $(f_n)$ is a Cauchy sequence of bounded functionals $f_n:V\to k$, take any $x$ and prove $f_n(x)$ is a Cauchy sequence in $k$. I'd take you assume your base field is complete, for example $k=\Bbb R$.

For future students, here is a more general result, taken from

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.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.