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Show that if $||\cdot||_1$, $||\cdot||_2$ are equivalent norms then $(V,||\cdot||_1)$ is a banach space iff $(V,||\cdot||_2)$ is. I really didn't get it. Of course both spaces are normed spaces but there are two things I don't understand:

  1. Why is it so important to look at "Cauchy sequences"? Why can't I look at converging sequences? Aren't those equivalent?

  2. If those are equivalent, ones a sequence converges to a limit (which is in the space) should it be different in the other "space", that is $(V,||\cdot||_2)?

I really didn't understand what I need to do here. I would truly appreciate your help.

Meitar
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  • A complete metric space is one in which all Cauchy sequences converge. So they are equivalent iff the norm results in a banach space. – Matthew Leingang Oct 30 '15 at 14:09
  • But the question was stated in a very odd way, as if the norm has a saying in whether or not the space is Banach...Is it really rational? – Meitar Oct 30 '15 at 14:12
  • We usually say “let $V$ be a Banach space,” but completeness is a property of the metric. So we really mean “let $(V,\left\Vert\cdot\right\Vert)$ be a Banach space.” – Matthew Leingang Oct 30 '15 at 14:13
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    Of course the norm has a saying in whether the space is a Banach space. Consider $\ell^1$. That's a Banach space with the usual norm. The norm $||x||_2=\sqrt{\sum|x_j|^2}$ is another norm on $\ell^1$, with respect to which it is not a Banach space. – David C. Ullrich Oct 30 '15 at 14:14
  • So I have to use only Cauchy? It has nothing to do with convergence of a sequence(which I believed to idependent )? – Meitar Oct 30 '15 at 14:16
  • A Banach space $V$ is a complete topological vector space (on a subfield $\mathbb{K}$ of $\mathbb{C}$) equipped with a norm (and the topology resulting from it). Topological vector space means that $V\times V\to V:(u,v)\mapsto u+v$ and $\mathbb{K}\times V:(\lambda,u)\mapsto \lambda u$ are continuous applications. Continuity explicitly depends on the topology, i.e. on the norm. Completeness means that any Cauchy sequence is convergent, which also depends on the norm. – MoebiusCorzer Oct 30 '15 at 14:17
  • A normed space is Banach if every Cauchy sequence converges to a limit in the space. Since V is Banach with respect to norm_1 you just have to show that a Cauchy sequence with norm _1 converges to the same limit w.r.t norm_2. (which isn't difficult). – Tom Collinge Oct 30 '15 at 14:19
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    It is important to look at Cauchy sequences because, in some spaces, there are Cauchy sequences that do not converge. In complete spaces (i.e. Banach spaces), all Cauchy sequences converge, which means that the two concepts happen to be equivalent. – Ben Grossmann Oct 30 '15 at 14:29
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    You have to use Cauchy because of the definition of the word "complete"! – David C. Ullrich Oct 30 '15 at 14:33

2 Answers2

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It suffices to show that

  1. a sequence $\left\{v_n\right\}$ is Cauchy in $\left\Vert\cdot\right\Vert_1$ iff it is Cauchy in $\left\Vert\cdot\right\Vert_2$.

  2. a sequence $\left\{v_n\right\}$ converges in $\left\Vert\cdot\right\Vert_1$ iff it converges in $\left\Vert\cdot\right\Vert_2$.

Why is this sufficient? Because then the argument goes like this: Suppose $\left\Vert\cdot\right\Vert_1$ is complete; that is, $V$ is a Banach space in the $\left\Vert\cdot\right\Vert_1$ metric. Let $\left\{v_n\right\}$ be a Cauchy sequence in $\left\Vert\cdot\right\Vert_2$. By (1.) the sequence is Cauchy in $\left\Vert\cdot\right\Vert_1$. Since $\left\Vert\cdot\right\Vert_1$ is complete, the sequence converges in $\left\Vert\cdot\right\Vert_1$, and by (2.) the sequence converges in $\left\Vert\cdot\right\Vert_2$. Therefore $\left\Vert\cdot\right\Vert_2$ is complete.

Matthew Leingang
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Because completeness means that all Cauchy sequences are convergent. Thus, to prove completeness obviously you have to take a Cauchy sequence and prove that it is convergent. This is so obvious that it's hard to understand your difficulty. Maybe you are used to Euclidean spaces, which are always complete, so there's nothing to prove in Euclidian space; but in infinite dimensional spaces you have to prove it.

user285394
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    “This is so obvious that it's hard to understand your difficulty” comes off as condescending to the OP. Indeed, some things are “obvious” only after they have been pondered for a while. See http://math.stackexchange.com/questions/151782/when-is-something-obvious – Matthew Leingang Oct 30 '15 at 14:27
  • I would expect people who find things obvious to find the concept of relativity obvious. – Meitar Oct 30 '15 at 14:40