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.

  • 1
    $\begingroup$ About Cauchy's inequality. Don't be too abstract. A version of that inequality is the one which follows: $$\tag{1}\sum_{i=1}^n x_i y_i \le \sqrt{\sum_{i=1}^n x_i^2}\sqrt{\sum_{i=1}^n y_i}, $$ and it holds for any $x_i, y_i\ge 0$. So you only have to rewrite (1) with $x_i=\lVert a_i\rVert,\, y_i=\lVert \phi_i\rVert.$ That's it, no need to worry about inner products or other abstract machinery. $\endgroup$ – Giuseppe Negro Jan 25 '13 at 18:29
  • 1
    $\begingroup$ Also $||\sum_i^n a_i\phi_i||\leq \sum^n_i|a_i|||\phi_i||$ is repeated use of the triangle inequality that $||a+b||\leq ||a||+||b||$. It is an inequality as we change everything between one set $||\cdot ||$ to between $n$ sets by that triangle inequality. $\endgroup$ – user45150 Jan 25 '13 at 18:31
  • $\begingroup$ of course! for some reason I was missing the step $||\sum_i^n a_i\phi _i||\leq \sum_i^n||a_i\phi _i||$ and was just thinking of these as equal! Thank you, sometimes one really messes up ones mind on stupid things! $\endgroup$ – harajm Jan 25 '13 at 18:50

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 $x_i,y_i \in \mathbb{R}$.


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