Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:

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.

share|cite|improve this question
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. – Giuseppe Negro Jan 25 '13 at 18:29
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. – user45150 Jan 25 '13 at 18:31
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! – harajm Jan 25 '13 at 18:50
up vote 4 down vote accepted

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.