I am working on a problem and i have pretty much a solution to every part ( total of three parts ). I would appreciate it, if you look at the solutions and let me know if they are acceptable. If not, how can i improve the solutions?
To the Problem:
Let $(V,\lvert\lvert.\rvert\rvert)$ be a normed vector space with $V \ne \{0\}$.
And let $(V^*,\lvert\lvert.\rvert\rvert_*)$ be a dual space with : $V^*:=(A:V \to \mathbb R;\ \text{$A$ is a linear and continuous functional})$
(a) Consider a Cauchy_sequence $(A_n)_{n\in \mathbb N}\in V^*$. Show that it exists $\lim_{n \to \infty} A_n(z),$ for all $z\in V$
(b) Let $A(z) := \lim_{n \to \infty} A_n(z)$ Show that A is linear and bounded.
(c) Show that for every $\epsilon > 0$ exists a $N \in \mathbb N$ so that: $$\lvert A(x)-A_n(x)\rvert \le \epsilon.\lvert\lvert x\rvert\rvert$$ for all $n \ge N$ and all $x \in V$
MY SOLUTIONS:
for (a) what would be an appropriate Sequence to work with?
for (b) i did similar to a to show that A(x) is linear. But i can not show that A(x) is bounded! I approach it similar to (a) but at the end i have $C = \lim_{n \to \infty} 1/n = 0$ which is against $C > 0$.
What is the best way to approach (c) using the fact that A_n is a Cauchy-set?
Thank you for your Answers.
(a) Given a Cauchy sequence $A_n \in V^*$ and a point $z \in V$, look at the sequence $A_n(z) \in \mathbb{R}$. Thanks to completeness of $\mathbb{R}$, in order to show that it's convergent, it's enough to show that it's a Cauchy sequence (in $\mathbb{R}$). What can you say about the difference $A_n(z)-A_m(z)$, based on the assumption (that $A_n \in V^*$ is a Cauchy sequence)?
(b) Linearity of $A$ is the easiest part of the whole exercise. As for boundedness, it follows from the Cauchy criterion that $A_n$ is bounded. That is, there is $C \ge 0$ such that $\| A_n \| \le C$ for each $n$. This tells you how big $A_n(z)$ can be, and in the limit - how big $A(z)$ can be.
(c) From the Cauchy criterion, you should be able to check that $$ \forall \, \varepsilon>0 \ \exists \, N \in \mathbb{N} \ \forall \, m,n \ge N, \, x \in V \qquad |A_m(x)-A_n(x)| \le \varepsilon \| x \|. $$ Just take the limit $m \to \infty$.