# The dual vector space is always complete. [duplicate]

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?

## marked as duplicate by Arnaud D., Rolf Hoyer, José Carlos Santos, Ross Millikan, астон вілла олоф мэллбэргNov 22 '17 at 23:38

• I assume $N^*$ is the space of continuous linear functionals, otherwise, I do not believe the result is true. As for your question, addition and scalar multiplication are continuous. – Paul Sinclair Oct 1 '15 at 13:44
• Yes, $N^*$ is the set of all continuous linear functionals. But I do not see how the latter part of your comment is an answer. Could you please give a more explicit answer? – DK26 Oct 1 '15 at 13:53
• Show that $\{f_n\}$ is Cauchy implies $\{f_n(x)\}$ converges for all $x$ and let $f(x)$ be the point-wise limit. Use continuity of addition and scalar multiplication to show that $af_n(x)+bf_n(y)$ converges to $af(x) + bf(y)$. And show that continuity of $f_n$ for sufficiently high $n$ implies continuity of $f$. – Paul Sinclair Oct 1 '15 at 14:01
• Does this mean that all reflexive vector spaces are Banach iff the base field is complete? – DK26 Oct 1 '15 at 14:12
• Evidently so!-- – Paul Sinclair Oct 1 '15 at 19:57

For $0\neq x\in N$ and some $\delta>0$ choose $\varepsilon=\frac{\delta}{||x||}$. So $|f_n(x)-f_m(x)|<||fn-f_m||||x||=\varepsilon||x||=\delta$ 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 $||.||$ is the operator norm) it is bounded, thus a functional.

• Why is $f$ linear? – DK26 Oct 1 '15 at 14:02
• I tried to work out the details in the answer below – nelv Oct 1 '15 at 14:05
• because $f(rx+y)=\lim_{n\rightarrow \infty} f_n(rx+y)=\lim_{n\rightarrow \infty}rf_n(x)+f_n(y)=r\lim_{n\rightarrow \infty}fn(x)+\lim_{n\rightarrow \infty}f_n(y)=rf(x)+f(y)$ – Peter Melech Oct 1 '15 at 14:05
• @nelv thanks i was not that fast. – Peter Melech Oct 1 '15 at 14:06
• $||f_n-f||\rightarrow 0$ is still missing, – Peter Melech Oct 1 '15 at 14:07

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 $$+: K^2\to K\ :\ (x,y)\mapsto x+y$$ and $$\lambda\cdot: K\to K\ :\ x\mapsto \lambda\cdot x,\ \forall\ \lambda\in K$$ are continuous. Then: $$f(x+y) = lim_{n\to\infty}f_n(x+y)=lim_{n\to\infty}[+(f_n(x),f_n(y))]= \\ =+(lim_{n\to\infty}f_n(x),lim_{n\to\infty}f_n(y)) = +(f(x),f(y))=f(x)+f(y)$$ where I used to continuity to swap the map + with the limit. The proof of scalar multiplication is completely analogous.

I hope this helps. :)

• This tells me why $f$ is linear, thank you. I have boundedness/continuity from the other answers. – DK26 Oct 1 '15 at 14:07

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 $f_n \to f$


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 $f$.