Any two norms equivalent on a finite dimensional norm linear space.

Question

I am trying to understand the proof that every two norms on a finite dimensional NLS are equivalent.

I am working with this proof I found on the web: http://www.math.colostate.edu/~yzhou/course/math560_fall2011/norm_equiv.pdf

My first question is, do we always assume that the space is over $\mathbb{R}$ or $\mathbb{C}$ and not an arbitrary field?

Secondly, let $\Phi =\{\phi _1, \phi _2...,\phi _n\}$ be a basis for $H$, for $x\in H$ we have $x=\sum_{i}^na_{i}\phi _i$. I guess we are assuming $a_i\in \mathbb{R}$?

We then define $p(x)= \sqrt{\sum_{i}^na_{i}^2}$ We prove that $p(x)$ is a norm.

Now we want to show that every norm on $H$ is equivalent to $p(x)$. I wonder about the "easy" inequality, that it exists $M$ such that for $||.||$ an arbitrary norm $||x||\leq Mp(x)$. To prove this it is stated in every proof I have found that $||\sum_{i}^na_{i}\phi _i||\leq \sum_{i}^n||a_i||*||\phi _i||$, I don't understand this, $a_i$ is a scalar right, by the norm axiom $||a_i \phi _i ||=|a_i|*||\phi _i||$, hence strict equality?

Furthermore in the next sentence Cauchys inequality is used: $\sum_{i}^n||a_i||*||\phi _i||\leq \sqrt{\sum_{i}^n||a_i||}*\sqrt{\sum_{i}^n||\phi_i||}$

How can they assume that this arbitrary norm is associated to an innerproduct???

I wonder about the above two questions specifically.

Answer 1

1) The norms are equivalent in every vector spaces over a complete valued field, but the proof is more difficult. For example, see Topological Vector Spaces by Bourbaki.

2) For the inequality $\displaystyle \left\|\sum\limits_{i=1}^n a_i\phi_i\right\| \leq \sum\limits_{i=1}^n \|a_i\| \cdot \|\phi_i\|$, just apply the triangular inequality and homogeneity.

3) Then, Cauchy-Schwarz inequality, as it is used here, is only $\displaystyle \sum\limits_{i=1}^n x_iy_i \leq \sqrt{ \sum\limits_{i=1}^n x_i^2} \cdot \sqrt{ \sum\limits_{i=1}^n y_i^2}$ for any $x_i,y_i \in \mathbb{R}$.