2

Let X be a linear space, why is the dual space of X a Banach space? Can anyone please explain?

user1559897
  • 1,807
  • 2
    If $Y$ is Banach and $X$ is normed, the space of bounded linear functionals $L(X,Y)$ is always Banach. – Pedro Sep 09 '16 at 01:06
  • 2
    It is not so, in general. So not "seeing why it is" is very reasonable. Still, ... context? – paul garrett Sep 09 '16 at 01:06
  • 1
    Because if $\phi_n$ is a Cauchy sequence in $X^{}$ then $\phi_n (x)$ is a Cauchy sequence in the base field, which is complete.You can then prove that $\phi_n$ converges in $X^{}$ – Matematleta Sep 09 '16 at 01:08
  • @paulgarrett It is, because $\mathbb R$ is Banach. =) – Pedro Sep 09 '16 at 01:11
  • 1
    If $X$ does not come with a norm what do you mean by the dual? Which norm there? – Jochen Sep 09 '16 at 11:11

1 Answers1

2

I'll show you how to set it up. If you need more assistance, let me know and I will edit.

Let $\{f_n\}$ be a Cauchy sequence in $X^*$ (the dual space of $X$). Then for each $x\in X$, $\{f_n(x)\}$ is a Cauchy sequence in the scalar field, and hence convergent (because both $\mathbb{R}$ and $\mathbb{C}$ are complete metric spaces). Now define the function $f:X\to\mathbb{C}$ by $$ f(x)=\lim_{n\to\infty}f_n(x).$$

Now we need to show two things.

(i) $f\in X^*$ (i.e. $f$ is linear and bounded)

(ii) $f_n\to f$ in norm (i.e. $\|f_n-f\|\to0$ as $n\to\infty$).

Aweygan
  • 23,232