Fréchet Space (Functional Analysis) is Metric Space

Theorem
Let $\left({\R^\omega, d}\right)$ be the Fréchet space on $\R^\omega$.

Then $\left({\R^\omega, d}\right)$ 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:
 * $\forall x, y \in \R^\omega: d \left({x, y}\right) = \displaystyle \sum_{i \mathop \in \N} \dfrac {2^{-i} \left\lvert{x_i - y_i}\right\rvert} {1 + \left\lvert{x_i - y_i}\right\rvert}$

where $x := \left\langle{x_i}\right\rangle_{i \mathop \in \N} = \left({x_0, x_1, x_2, \ldots,}\right)$ and $y := \left\langle{y_i}\right\rangle_{i \mathop \in \N} = \left({y_0, y_1, y_2, \ldots,}\right)$ denote arbitrary elements of $\R^\omega$.

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

Proof of $M1$
So axiom $M1$ holds for $d$.

Proof of $M3$
So axiom $M3$ holds for $d$.

Proof of $M4$
Thus for such a $k \in \N$:

It has been established during the course of demonstrating compliance with axiom $M1$ (and is in any case trivially obvious) that:
 * $x_j = y_j \implies \dfrac {2^{-j} \left\lvert{x_j - x_j}\right\rvert} {1 + \left\lvert{x_j - x_j}\right\rvert} = 0$

Thus $\displaystyle \sum_{i \mathop \in \N} \dfrac {2^{-i} \left\lvert{x_i - x_i}\right\rvert} {1 + \left\lvert{x_i - x_i}\right\rvert}$ contains at least one term:
 * $\dfrac {2^{-k} \left\lvert{x_k - x_k}\right\rvert} {1 + \left\lvert{x_k - x_k}\right\rvert} > 0$

and any number of other terms:


 * $\dfrac {2^{-j} \left\lvert{x_j - x_j}\right\rvert} {1 + \left\lvert{x_j - x_j}\right\rvert} = 0$

Hence:
 * $\displaystyle \sum_{i \mathop \in \N} \dfrac {2^{-i} \left\lvert{x_i - x_i}\right\rvert} {1 + \left\lvert{x_i - x_i}\right\rvert} > 0$

for $x \ne y$.

So axiom $M4$ holds for $d$.

Proof of $M2$
Consider the function $f (a) = a / (1+a)$, $a \geq 0$. We want to show it satisfies


 * $f (a) \leq f(a') \quad \text{ for } \; a \geq 0 \; \text{ and } \; a' \geq a$.

This very first inequality follows since $f (a) = 1 - 1/ (1+a)$ with $1/(1+a)$ decreasing as $a$ increases. We also want to show it satisfies


 * $f (a) \leq f(a+b) \leq f(a) + f(b) \quad \text{ for } \; a,b >0$.

The first inequality here follows from the above inequality. For the second inequality here we note $f(a) / a = 1/(1+a)$ gives


 * $f (a)/a \geq f(a+b)/(a+b), \quad f(b)/b \geq f(a+b)/(a+b) \text{ for } \; a,b >0$

from which we obtain $f(a) + f(b) \geq f(a+b) (a+b) / (a+b) = f (a+b)$ (the inequality is immediately verified in cases where $a$ or $b$ is $0$).

We now use these inequalities to prove axiom $M2$

So axiom $M2$ holds for $d$.

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