Take the 2-minute tour ×
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.

I'm working through Rudin's Functional Analysis, and I am confused by a step in his proof for Theorem 1.24, which states that if X is a topological vector space with a countable local base, then there is a metric d on X such that a) d is compatible with the topology of X b) the open balls are balanced c) d is invariant and, if X is locally convex, then all open balls are convex.

Here is a slimmed down version of his proof:

By his theorem 1.14, X has a balanced local base $\{V_n\}$ such that $V_{n+1} + V_{n+1} + V_{n+1} + V_{n+1} \subset V_n$.

D is the set of all rational numbers r of the form $\Sigma_{n=1}^{\infty} c_n(r) 2^{-n}$, where each of the digits $c_i(r)$ is 0 or 1 and only finitely many are 1. Define A(r) = X if $r \geq 1$. For any r in D, define A(r) = $\Sigma c_n(r)V_n$. (Note each is a finite sum). Define $f(x) = \inf \{r : x \in A(r)\}$ for $x \in X$ and d(x,y) = f(x - y). Rudin proves an inclusion $A(r) + A(s) \subset A(r + s)$ and uses it to show that $\{A(r)\}$ is totally ordered by set inclusion and that $f(x + y) \leq f(x) + f(y)$.

As each A(r) is balanced, $f(x) = f(-x)$. $f(0) = 0$. If x$\not = 0$, then $x \not \in V_n = A(2^{-n})$ for some n, so $f(x) \geq 2^{-n} > 0$.

I don't understand how he then concludes that $d(x,y) = f(x -y)$ - "these properties of f show that d(x,y) defines a translation invariant metric d on X$."

Specifically, it is obvious that if z = 0, then $d(x + z, y + z) = d(x,y)$. But to show $d(x + z, y + z) \leq d(x,y)$, I have tried this: $d(x + z, y + z) = f( (x + z) - (y+ z)) \leq f(x) + f(y) + 2f(z)$, which overestimates $f(x) + f(y)$ since $f(z)> 0$ and thus shows diddly-squat.

I know I'm probably being profoundly thick, but I am frustrated and would like to move on. Thoughts?

Thanks in advance. More thanks to follow.

share|improve this question
    
How do we know here that $ \forall x \in X, \exists r, s.t. x \in A(r)$? –  user93171 Sep 5 '13 at 0:10
    
He defines $A(r) = X$ if $r \geq 1$. –  AreaMan Sep 6 '13 at 14:10

1 Answer 1

up vote 5 down vote accepted

Your question: How he concludes $d(x,y)=f(x−y)$? and how $d$ is translation invariant.? Addressing to this:

In the proof, he is not concluding $d(x,y)=f(x−y)$ but he is defining $d$ in this way. And he proves that $d$ is metric. In the proof he proves $f(x+y)\leq f(x)+f(y)$. This proves for any $z$, $d(x,z)\leq d(x,y)+d(y,z)$ and then he proves $d(x,x)=0$ if and only if $x=0$. All this shows that $d$ is metric. Now translation invariant follows by definition of $d$ itself.

$$d(x+z, y+z)= f((x+z)-(y+z))= f(x-y)= d(x,y)$$

share|improve this answer
1  
Thanks, you've made it much clearer. I was confused because I thought that in a topological vector space the vector addition operations didn't commute (but of course it works out, because $z + x = x + z, so (x + z) - (y + z) = (x + z) - (z + y) = x + (z - z) - y = x - y$) and so I was looking for those facts about f to show that the equation you've left in that final line holds, rather than to show that d is a metric. Thanks very much, it's clear now. –  AreaMan Jan 8 '13 at 7:46

Your Answer

 
discard

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.