showing a sequence is Cauchy in a normed space

Question

Suppose I have a sequence in a normed space. In this space, the Cauchy-ness and convergence of the sequence is determined completely by the specific norm defined on the space, correct? So when showing that a given sequence is Cauchy on this normed space, one would not be able to use any information about the sequence from an outside context (for example, that the sequence convergences under the standard absolute value norm in $\mathbb{R}$)?

As you can see, I am a little confused about non-standard norms on vector spaces. If we have one, is that the ONLY norm we can have there? Or can we have many different norms, including the standard one, on the space simultaneously? I hope that makes sense.

Answer 1

Most vector spaces can be given many norms. However, in the case of finite dimensional vector spaces, all possible norms are equivalent: i.e. if $|\cdot|$ and $||\cdot||$ are two norms on $V$, there exist constants $a,b>0$ such that for all $v\in V$, we have $a|v|\le ||v||\le b|v|$. If two norms are equivalent, then you should be able to check that a sequence converges in one iff it converges in the other.

Answer 2

Here is a proof that all norms on finite dimensional vector spaces are equivalent, this is meant to supplement helloworld's answer.

Let X be a finite dimensional vector space. We prove that every norm on X is equivalent to the sup norm. First, recall, that every finite dimensional vector space as a basis which we denote by $\{e_i\}_{i=1}^{n}$. Let $x \in X$ be arbitrary then $x = \sum\limits_{i=1}^{n}{c_i e_i}$, where the $\{c_i\}$ are in the underlying field. Let $||\cdot ||$ be arbitrary. Then we have that $||x||=||\sum\limits_{i=1}^{n}{c_i e_i}|| \leq \sum\limits_{i=1}^{n}{||c_i e_i||} \leq n \max_{i \in \{1,..,n\}}{||e_i||} ||x||_{\infty}$. \ To give the second bound, we first prove that all norms are continuous. Let $\epsilon > 0$ and let $x,y \in X$ where X is a linear space. Note that $|||x|| -||y||| \leq ||x-y||$ and this select $\delta = \epsilon$ and we have continuity. Moreoever, note that $\forall x \in X-\{0\}$ that $x/||x|| \in S^1$. Note that $S^1=\{x \in X: ||x||=1\}$ is a closed and bounded set and thus is compact. Thus, we get that $||\frac{x}{||x||}||_{\infty} \leq M$ since our norm is continuous on a compact set and is bounded. Then we get that $||x||_{\infty}/M \leq ||x||$. Thus, we have that $\frac{1}{M} ||x||_{\infty} \leq ||x|| \leq n \max_{i \in \{1,..,n\}}{||e_i||} ||x||_{\infty}$. Now, if $x=0$ then we trivially get the result. Thus, we have shown equivalence in a point-wise fashion and we have all norms are equivalent on finite dimensional vector spaces.

Caution: This result does not hold in infinite dimensional normed spaces. So to answer your question in general: Cauchyness or convergence of a specific is related only to the norm it is converging in. Now, if you have equivalent norms then it easy to see as Helloworld has exhibited that it will converge then in both spaces.