10
$\begingroup$

I am trying to prove that every vector space $X$ has a norm. I have some silly questions, but it's better to ask them now instead of later. I think I'm having a bit of trouble getting intuition about basis in infinite dimensional spaces.

Fix a Hamel basis ${\cal B} = ({\bf e}_i)_{i \in I}$. Then for all ${\bf x} \in X$, we write: $${\bf x} = \sum_{i \in F}a_i{\bf e}_i,$$ for some $F \subset I$ finite. I understand that this combination is unique in the sense that: $$\sum_{i \in F_1}a_i{\bf e}_i = \sum_{i \in F_2}b_i{\bf e}_i,$$ for some $F_1,F_2 \subset I$ finite, implies that $a_i = b_i$ for all $i \in F_1 \cap F_2$ and $a_i = 0$ for all $i \in F_1 \setminus F_2$, and $b_i = 0$ for all $i \in F_2 \setminus F_1$.

Does this mean that $F_1 = F_2$?

Assuming that yes, although I'm not sure, the idea would be to define some kind of max norm, which would be well-defined: $$\|{\bf x}\| = \max\{|a_i| \mid i \in F \}.$$ This idea seems good, it even showed up in another answer. The properties $\|{\bf x}\| \geq 0$ for all ${\bf x}\in X$, $\|{\bf x}\| = 0 \implies {\bf x}=0$ and $\|\lambda{\bf x}\| = |\lambda|\|{\bf x}\|$ are all clear. I'm having trouble getting the triangle inequality. How to make a sum here is a bit confuse to me. If ${\bf x},{\bf y}\in X$, then there are $F_1,F_2 \subset I$ finite such that: $${\bf x} = \sum_{i \in F_1}a_i{\bf e}_i, \quad \text{and}\quad {\bf y}=\sum_{i\in F_2}b_i{\bf e}_i,$$ so that $${\bf x}+{\bf y} = \sum_{i \in F_1 \setminus F_2}x_i{\bf e}_i + \sum_{i\in F_1 \cap F_2}(a_i+b_i){\bf e}_i + \sum_{i \in F_2 \setminus F_1}b_i{\bf e}_i.$$ I wanted to write this as $\sum_{i \in \text{ something}}c_i{\bf e}_i,$ but the only thing I could think of was: $$\sum_{i \in F_1 \cup F_2}c_i{\bf e}_i, \quad c_i = \begin{cases} a_i, \text{ if }i \in F_1 \setminus F_2 \\ a_i+b_i, \text{ if }i \in F_1 \cap F_2 \\ b_i, \text{ if }i \in F_2 \setminus F_1\end{cases}$$

But again:

Is this combination unique in the sense that the only possible combination for the vector ${\bf x}+{\bf y}$ will be indexed by $F_1\cup F_2$?

I think I am overcomplicating things. I can get the triangle inequality with this, it seems, but things don't look well-defined enough for me. Can someone address these questions and give me a small explanation about it? Thanks.


Edit: to confirm what I understood from gerw's answer: since the combination is unique, if I write $${\bf x}=∑_{i∈F_1}a_i{\bf e}_i=∑_{i∈F_2}b_i{\bf e}_i,$$ for $F_1,F_2⊂I$, finite sets, then: $\max\{|a_i|∣i∈F_1\}=\max\{|b_i|∣i∈F_2\}$, and this ensures that $\|\cdot\|$ is well defined, right?

$\endgroup$
4
$\begingroup$
  1. No, you do not necessarily have $F_1 = F_2$. But as you have already shown the coefficiencts of the indices in $(F_1 \setminus F_2) \cup (F_2 \setminus F_1)$ are zero.

  2. This combination is unique, up to zero coefficients, cf. 1.

$\endgroup$
  • $\begingroup$ Thanks for the answer. Can you please confirm if I understood everything ok? Since the combination is unique, if I write $${\bf x} = \sum_{i \in F_1}a_i{\bf e}_i = \sum_{i \in F_2}b_i{\bf e}_i,$$ for $F_1,F_2 \subset I$, finite sets, then: $$\max\{ |a_i| \mid i \in F_1 \} = \max\{ |b_i| \mid i \in F_2 \},$$ and this ensures that $\|\cdot\|$ is well defined, right? $\endgroup$ – Ivo Terek May 29 '15 at 19:00
  • $\begingroup$ Yes, you are right. $\endgroup$ – gerw May 29 '15 at 19:24
  • $\begingroup$ Ok. Thanks again! $\endgroup$ – Ivo Terek May 29 '15 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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