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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.

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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
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.

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

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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

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