The dual vector space is always complete.

Question

If $N$ is a normed a linear space, then its dual vector space $N^*$ is always complete.

Attempt: Let $\{f_n\}$ be a Cauchy sequence in $N^*$. Then, for some $\varepsilon > 0$, there exists $m,n \in \mathbb{N}$ such that $\|f_n - f_m \| < \varepsilon$. The way to show that the limit lies in $N^*$ would be to show that $f_n(x)$ converges in $K = \mathbb{C}$ or $\mathbb{R}$. How do I show that the limit of linear functionals is still a linear functional?

Answer 1

For $0\neq x\in N$ and some $\delta>0$ choose $\varepsilon=\frac{\delta}{\| x \|}$. So $|f_n(x)-f_m(x)|< \|f_n-f_m\| \|x\|=\varepsilon\|x\|=\delta$ for sufficiently large $n,m$. So for any $x\in N$ (trivially for $x=0$)the sequence $\{f_n(x)\}$ is Cauchy in $\mathbb{C}$ and thus has a limit. Then define $f$ by $f(x):=\lim_{n\rightarrow \infty}f_n(x)$. It is clear that $f$ is linear and by

$\|f\| \leq \| f-f_n \|+ \|f_n \|<\varepsilon + \| f_n \| <\infty$

(where $\| \cdot \|$ is the operator norm) it is bounded, thus a functional.

Answer 2

observe that $||f_m-f_n||\leqslant \epsilon$ implies $\{f_n(x)\}_{1\leqslant n}$ is a cauchy sequence in $\mathbb{C}$ and hence convergent so now define a function $f$ given by $f(x)={lim}_{n \to \infty}f_n (x)$ and show that $f \in N^* $ and $f_n \to f$

Answer 3

As remarked in the comments, to show that your $f$ is linear you can use the fact that addition and scalar multiplication and continuous. Formally, the maps \begin{equation} +: K^2\to K\ :\ (x,y)\mapsto x+y \end{equation} and \begin{equation} \lambda\cdot: K\to K\ :\ x\mapsto \lambda\cdot x,\ \forall\ \lambda\in K \end{equation} are continuous. Then: \begin{equation} f(x+y) = lim_{n\to\infty}f_n(x+y)=lim_{n\to\infty}[+(f_n(x),f_n(y))]= \\ =+(lim_{n\to\infty}f_n(x),lim_{n\to\infty}f_n(y)) = +(f(x),f(y))=f(x)+f(y) \end{equation} where I used to continuity to swap the map + with the limit. The proof of scalar multiplication is completely analogous.

I hope this helps. :)

Answer 4

$\newcommand{\norm}[1]{\left \lVert #1 \right \rVert}$ $\newcommand{\abs}[1]{\lvert #1 \rvert}$

Since $\{f_n \}_{n \in \mathbb{N}}$ is Cauchy in $(W^*, \norm{\cdot}_*)$ , we have for each $x \in (W, \norm{\cdot})$ $$ |f_n(x)-f_m(x)| \le \norm{f_n-f_m}_* \norm{x} \to 0 \quad \text{ as } m, n \to \infty. $$ This means $\{ f_n(x) \}_{n \in \mathbb{N}}$ is Cauchy in $\mathbb{R}$ for any $x \in W$. Because $(\mathbb{R}, \abs{\cdot})$ is complete, we can define $$ f(x) = \lim_{n \to \infty} f_n(x) \quad \forall x \in W. $$ Clearly, $f$ is a linear functional. It remains to show $f$ is continuous (or, equivalently, bounded.)

Note that any Cauchy sequence is bounded under the same norm. We then have some $M < \infty$ such that
$$ \norm{f_n}_* \le M \quad \forall n \in \mathbb N. \tag{$\star$} $$

Since $\abs{\cdot}$ is continuous in $(\mathbb R,\abs{\cdot})$, we have by ($\star$) that $$ \abs{f(x)} = \abs{\lim_{n \to \infty} f_n(x)} = \lim_{n \to \infty} \abs{f_n(x)} \le M\norm{x} \quad \forall x \in W, $$ proving continuity of $f$.