Theorem 1.24 Rudin, Functional analysis (first half)

Question

Reading through theorem 1.24 of Rudin functional analysis. I'm struggling to understand a bit the construction of the metric, and related properties. Rudin defines $D$ as the set of all rational of the form

$$ r = \sum_{n=1}^{+\infty} c_n(r)2^{-r} $$

where $c_i(r)$ is either $0$ or $1$ with finite many $1$. Now after defining

$$ A(r) = c_1(r) V_1 + c_2(r)V_2 + \ldots $$

we have the definition of the functional

$$ f(x) = \inf \left\{r : x \in A(r) \right\}\;\;,x \in X $$

(Here $X$ is a t.v.s. with a countable local base), it firstly assumes

$$ A(r)+A(s)\subset A(r+s) $$

before actually proving it, it is also easy to proove

$$ A(r) < A(t) \Leftrightarrow r < t $$ and now the first claim that puzzles me, I'm sure it is silly but still I struggle to understand. He proves

$$ f(x+y) \leq f(x)+f(y), $$ And the first thing said is

We may, of course, assume the right side is $< 1$

My first question is "why?". Right after

Fix $\epsilon>0$. There exists $r$ and $s$ in $D$ such that $$ f(x)<r,f(y)<s, r+s<f(x)+f(y)+\epsilon$$

I know this might sound silly now but I can understand why I can find $r,s$ such that $f(x)<r$ and $f(y)<s$, which is independent from $\epsilon$ (this follows from the definition of $f$) but why the can I also have $r+s<f(x)+f(y)+\epsilon$?

And later it is stated

If $x \neq 0$ then $x \notin V_n = A(2^{-n})$ for some $n$

While I have the intuition for this I struggle to prove this small step. And finally it is concluded the translation invariance of the metric $f$, I can get the "metric" but not the invariance....

Can you please clarify?

Answer 1

We can assume that the right hand side is $< 1$ since otherwise the inequality is trivially satisfied because $f(z) \leqslant 1$ for all $z$. So if $f(x) + f(y) \geqslant 1$, we have

$$f(x+y) \leqslant 1 \leqslant f(x) + f(y)\,.$$

Only for $f(x) + f(y) < 1$ is some work required to prove $f(x+y) \leqslant f(x) + f(y)$.

For the next problem, one cannot (usually) - and need not - find $r,s$ independent of $\epsilon$, the choice is made depending on $\epsilon$. Given any $\epsilon > 0$, by definition of $f$, namely $f(x) = \inf\:\{r : x \in A(r)\}$, we can find $r = r(\epsilon)$ such that $x \in A(r)$ and $r < f(x) + \epsilon/2$, and we can find $s = s(\epsilon)$ such that $y \in A(s)$ and $s < f(y) + \epsilon/2$. Then

$$x + y \in A(r) + A(s) \subset A(r+s)$$

whence $f(x+y) \leqslant r+s < f(x) + f(y) + \epsilon$. That $r$ and $s$ were chosen depending on $\epsilon$ is irrelevant, we just care about $f(x+y)$ and bounding that by $f(x) + f(y)$. Since $$f(x+y) < f(x) + f(y) + \epsilon$$ holds for every $\epsilon > 0$, the desired inequality $f(x+y) \leqslant f(x) + f(y)$ follows.

The family $\{ V_n : n \in \mathbb{N}\}$ is by assumption a neighbourhood base of $0$ in $X$. Since $X$ is $T_1$ (part of the definition of a TVS in Rudin's book), for every $x \neq 0$ there is a neighbourhood $W_x$ of $0$ that doesn't contain $x$. By definition of a neighbourhood base, there is an $n$ such that $V_n \subset W_x$. For this $n$ (and all larger) we thus have $x \notin V_n$.

A metric $\rho$ on $X$ is translation-invariant if for all $x,y,z \in X$ we have

$$\rho(x+z,y+z) = \rho(x,y)\,. \tag{$\ast$}$$

The metric $d$ is defined by $d(x,y) = f(x-y)$, and from $(x+z) - (y+z) = x-y$ we thus obtain the translation-invariance of $d$:

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

whatever $x,y,z \in X$ are.

Answer 2

For the second question, the result follows from the fact that $D$ is dense in $[0,1]$ and $\{A(r)\}$ is totally ordered by set inclusion.

To see that $D$ is dense in $[0,1]$, let $f(x)\in [0,1]$ be arbitrary. Let $c_1 = 0$ if $f(x) < 1/2$ and $c_1 = 1$ if $f(x)\ge 1/2$. Then, $f(x)\in[c_1/2,c_1/2+1/2]$. Let $c_2 = 0$ if $f(x)<c_1/2+1/4$ and $c_2 = 1$ if $f(x) \ge c_1/2+1/4$. Then, $f(x)\in[c_1/2+c_2/4,c_1/2+c_2/4+1/4]$. By continuing this procedure, we can make $r = c_1/2 + c_2/4 + \cdots + c_n/2^n$ arbitrarily close to $f(x)$.

Since $D$ is dense in $[0,1]$, for every $0\le f(x)<1$ and every $\epsilon>0$, we can always find $r\in D$ satisfying $f(x) < r < f(x) + \epsilon$. Since $f(x) = \inf\{r':x\in A(r')\}$ and $f(x) < r$, there exists $r'\in[f(x),r)$ such that $x\in A(r') \subset A(r)$.