Is the Euclidean norm the unique norm which turns $\mathbb{R}^n$ into a Banach space? 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.
 A: 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.
A: 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.
