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See the title. Essentially I'm wondering if the answer is no, does there exist two inequivalent notions of calculus on $\mathbb{R}^n$ obtained via the Frechet derivative for the two resulting (inequivalent) Banach Spaces.

Okazaki
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    If all norms over $\mathbb{R}^n$ are equivalent, then what is the inequivalence (of Banach spaces) you are referring to? – Ranc Aug 01 '17 at 13:34
  • @Ranc I mean in the case where the norms are not equivalent – Okazaki Aug 01 '17 at 13:37
  • But they are always equivalent. This is a standard result for $\mathbb{R}^n$. – Ranc Aug 01 '17 at 13:38
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    @Ranc I realise now you're referring to an actual result "all norms on $\mathbb{R}^n$ are equivalent", sorry, I was actually unaware of this fact (revising Banach spaces after a very long time away from them) – Okazaki Aug 01 '17 at 13:38
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    In fact, we have even a slightly stronger result. Every finite dimensional normed space is a Banach space (this is can be proven using the fact that all norms on $\mathbb{R}^n$ are equivalent). – Severin Schraven Aug 01 '17 at 13:40
  • @Ranc If you rewrite the statement that all of the norms are equivalent as an answer, I will accept it. (Thank you for clarifying this for me btw) – Okazaki Aug 01 '17 at 13:41
  • @Okazaki it's ok :). Just wasn't sure if you needed this or something else. – Ranc Aug 01 '17 at 13:54
  • @SeverinSchraven GREAT point, Severin and people usually don't point this out in elementary analysis courses that use Banach spaces. – Mathemagician1234 Aug 30 '17 at 22:08

2 Answers2

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All norms on $\mathbb{R}^n$ (or any other finite-dimensional normed space) are equivalent, and so if a sequence converges or is a Cauchy sequence under one norm, so it does or is in any other. In particular, any norm on $\mathbb{R}^n$ turns it into a Banach space.

Reiner Martin
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Let $||.||_a$ be another norm in $\mathbb{R}^n$ .

We know that in $\mathbb{R}^n$ and in every finite dimensional space all the norms are equivalent, i.e:$$\exists C,D>0 :C||x||_a \leqslant ||x||_2 \leqslant D||x||_a, \forall x \in \mathbb{R}^n$$

Where $||.||_2$ is the euclideian norm.

Now let $\{v_1,v_2...v_n\}$ a basis of $\mathbb{R}^n$ and $x_m \in \mathbb{R}^n$ a Cauchy sequence with respect to the norm $||.||_a$. Note that $x_m=x_m^{(1)}v_1+...+x_m^{(n)}v_n$

We have that $\forall \epsilon>0, \exists m_0 \in \mathbb{N}$ such that $||x_s-x_l||_a< \epsilon, \forall s,l \geqslant m_0$

Using the inequality of the equivalence of norms you can see that $x_m$ is a cauchy sequence with respect to the euclideian norm and because of the completeness of $\mathbb{R}^n$ under the euclideian norm we have that $\exists x \in \mathbb{R}^n$ such that $||x_m-x||_2 \rightarrow 0$ and $x=x_1v_1+...+x_nv_n$

Again from the inequality of the equivalence of norms you can see that $||x_m-x||_a \rightarrow 0$.

Therefore $\mathbb{R}^n$ is complete with respcet to the norm $||.||_a$

Also you have to know that completeness is not preserved under homeomorphisms but it is preserved under isometries.

Marios Gretsas
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