Fréchet Space (Functional Analysis) is Metric Space

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Theorem

Let $\struct {\R^\omega, d}$ be the Fréchet space on $\R^\omega$.

Then $\struct {\R^\omega, d}$ is a metric space.


Proof

It is to be demonstrated that $d$ satisfies all the metric space axioms.

Recall from the definition of the Fréchet space that the distance function $d: \R^\omega \times \R^\omega \to \R$ is defined on $\R^\omega$ as:

$\forall x, y \in \R^\omega: \map d {x, y} = \ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - y_i} } {1 + \size {x_i - y_i} }$

where:

$x := \sequence {x_i}_{i \mathop \in \N} = \tuple {x_0, x_1, x_2, \ldots}$
$y := \sequence {y_i}_{i \mathop \in \N} = \tuple {y_0, y_1, y_2, \ldots}$

denote arbitrary elements of $\R^\omega$.


First it is confirmed that Fréchet Product Metric is Absolutely Convergent on arbitrary $x$ and $y$, as follows.


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First note that:

$\dfrac {\size {x_i - y_i} } {1 + \size {x_i - y_i} } = 1 - \dfrac 1 {1 + \size {x_i - y_i} }$

implies that we have:

$0 \le \dfrac {\size {x_i - y_i} } {1 + \size {x_i - y_i} } < 1$

By the Ratio Test we have that the series is absolutely convergent.

An important property of absolutely convergent series is that any rearrangement of the terms of the series is also convergent and equal to the original series.

This will be used in the proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality.


Proof of Metric Space Axiom $(\text M 1)$

\(\ds \map d {x, x}\) \(=\) \(\ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - x_i} } {1 + \size {x_i - x_i} }\) Definition of $d$
\(\ds \) \(=\) \(\ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size 0} {1 + \size 0}\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop \in \N} \dfrac 0 1\)
\(\ds \) \(=\) \(\ds 0\)

So Metric Space Axiom $(\text M 1)$ holds for $d$.

$\Box$


Proof of Metric Space Axiom $(\text M 3)$

\(\ds \map d {x, y}\) \(=\) \(\ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - y_i} } {1 + \size {x_i - y_i} }\) Definition of $d$
\(\ds \) \(=\) \(\ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {y_i - x_i} } {1 + \size {y_i - x_i} }\) Definition of Absolute Value
\(\ds \) \(=\) \(\ds \map d {y, x}\) Definition of $d$

So Metric Space Axiom $(\text M 3)$ holds for $d$.

$\Box$


Proof of Metric Space Axiom $(\text M 4)$

\(\ds x\) \(\ne\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds \exists k \in \N: \, \) \(\ds x_k\) \(\ne\) \(\ds y_k\)


Thus for such a $k \in \N$:

\(\ds x_k\) \(\ne\) \(\ds y_k\)
\(\ds \leadsto \ \ \) \(\ds \size {x_k - y_k}\) \(>\) \(\ds 0\) Definition of Absolute Value
\(\ds \leadsto \ \ \) \(\ds 2^{-i} \size {x_k - y_k}\) \(>\) \(\ds 0\)
\(\, \ds \land \, \) \(\ds 1 + \size {x_k - y_k}\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {2^{-k} \size {x_k - x_k} } {1 + \size {x_k - x_k} }\) \(>\) \(\ds 0\)


It has been established during the course of demonstrating compliance with Metric Space Axiom $(\text M 1)$ (and is in any case trivially obvious) that:

$x_j = y_j \implies \dfrac {2^{-j} \size {x_j - x_j} } {1 + \size {x_j - x_j} } = 0$


Thus $\ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - x_i} } {1 + \size {x_i - x_i} }$ contains at least one term:

$\dfrac {2^{-k} \size {x_k - x_k} } {1 + \size {x_k - x_k} } > 0$

and any number of other terms:

$\dfrac {2^{-j} \size {x_j - x_j} } {1 + \size {x_j - x_j} } = 0$


Hence:

$\ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - x_i} } {1 + \size {x_i - x_i} } > 0$

for $x \ne y$.


So Metric Space Axiom $(\text M 4)$ holds for $d$.

$\Box$


Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality

Consider the real function:

$\forall a \in \R_{\ge 0}: \map f a = \dfrac a {1 + a}$

We want to show $f$ satisfies:

$\map f a \le \map f {a'}$

for $a \ge 0$ and $a' \ge a$.


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The first inequality follows since:

$\map f a = 1 - \dfrac 1 {1 + a}$

with $\dfrac 1 {1 + a}$ decreasing as $a$ increases.

We also want to show it satisfies :

$\map f a \le \map f {a + b} \le \map f a + \map f b$

for $a, b > 0$.

The first inequality here follows from the above inequality.

For the second inequality here we note $\dfrac {\map f a} a = \dfrac 1 {1 + a}$ gives:

$\dfrac {\map f a} a \ge \dfrac {\map f {a + b} } {a + b}$
$\dfrac {\map f b} b \ge \dfrac {\map f {a + b} } {a + b}$

for $a, b > 0$

from which we obtain $\map f a + \map f b \ge \dfrac {\map f {a + b} } {a + b} = \map f {a + b}$ (the inequality is immediately verified in cases where $a$ or $b$ is $0$).

We now use these inequalities to prove Metric Space Axiom $(\text M 2)$: Triangle Inequality:

\(\ds \map d {x, z}\) \(=\) \(\ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - z_i} } {1 + \size {x_i - z_i} }\) Definition of $d$
\(\ds \) \(=\) \(\ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {\paren {x_i - y_i} + \paren {y_i - z_i} } } {1 + \size {\paren {x_i - y_i} + \paren {y_i - z_i} } }\)
\(\ds \) \(\le\) \(\ds \sum_{i \mathop \in \N} \dfrac {2^{-i} (\size {x_i - y_i} + \size {y_i - z_i} } {1 + \size {x_i - y_i} + \size {y_i - z_i} }\) as $\size {\paren {x_i - y_i} + \paren {y_i - z_i} } \le \size {x_i - y_i} + \size {y_i - z_i}$ and $f(a) \le f(a')$ for $a' \ge a$
\(\ds \) \(\le\) \(\ds \sum_{i \mathop \in \N} \paren {\dfrac {2^{-i} \size {x_i - y_i} } {1 + \size {x_i - y_i} } + \dfrac {2^{-i} \size {y_i - z_i} } {1 + \size {y_i - z_i} } }\) as $\map f {\paren {x_i - y_i} + \paren {y_i - z_i} } \le \map f {x_i - y_i} + \map f {y_i - z_i}$
\(\ds \) \(=\) \(\ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - y_i} } {1 + \size {x_i - y_i} } + \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {y_i - z_i} } {1 + \size {y_i - z_i} }\) Fréchet Product Metric is Absolutely Convergent
\(\ds \) \(=\) \(\ds \map d {x, y} + \map d {y, z}\) Definition of $d$


So Metric Space Axiom $(\text M 2)$: Triangle Inequality holds for $d$.

$\Box$


Thus $d$ satisfies all the metric space axioms and so is a metric.

$\blacksquare$

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