Hilbert Sequence Space is Metric Space
Theorem
Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{i \mathop \ge 0} {x_i}^2$ is convergent.
Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$.
Then $\ell^2$ is a metric space.
Proof 1
$\ell^2$ is a particular instance of the general $p$-sequence space $\ell^p$.
Hence $p$-Sequence Space of Real Sequences is Metric Space can be applied directly.
$\blacksquare$
Proof 2
By definition of the Hilbert sequence space on $\R$:
Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{i \mathop \ge 0} {x_i}^2$ is convergent.
Then $\ell^2 := \struct {A, d_2}$ where $d_2: A \times A: \to \R$ is the real-valued function defined as:
- $\ds \forall x = \sequence {x_i}, y = \sequence {y_i} \in A: \map {d_2} {x, y} := \paren {\sum_{k \mathop \ge 0} \paren {x_k - y_k}^2}^{\frac 1 2}$
From Convergence of Square of Linear Combination of Sequences whose Squares Converge we have that $\ds \sum_{k \mathop \ge 0} \paren {x_k - y_k}^2$ does actually converge.
Proof of Metric Space Axiom $(\text M 1)$
| \(\ds \map {d_2} {x, x}\) | \(=\) | \(\ds \paren {\sum_{k \mathop \ge 0} \paren {x_k - x_k}^2}^{\frac 1 2}\) | Definition of $d_2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\sum_{k \mathop \ge 0} 0^2}^{\frac 1 2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
So Metric Space Axiom $(\text M 1)$ holds for $d_2$.
$\Box$
Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality
Let $z = \sequence {z_i} \in A$.
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| \(\ds \map {d_2} {x, y} + \map {d_2} {y, z}\) | \(=\) | \(\ds \paren {\sum_{k \mathop \ge 0} \paren {x_k - y_k}^2}^{\frac 1 2} + \paren {\sum_{k \mathop \ge 0} \paren {y_k - z_k}^2}^{\frac 1 2}\) | Definition of $d_2$ | |||||||||||
| \(\ds \) | \(\ge\) | \(\ds \paren {\sum_{k \mathop \ge 0} \paren {x_k - z_k}^2}^{\frac 1 2}\) | Minkowski's Inequality for Sums: index $2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map {d_2} {x, z}\) | Definition of $d_2$ |
So Metric Space Axiom $(\text M 2)$: Triangle Inequality holds for $d_2$.
$\Box$
Proof of Metric Space Axiom $(\text M 3)$
| \(\ds \map {d_2} {x, y}\) | \(=\) | \(\ds \paren {\sum_{k \mathop \ge 0} \paren {x_k - y_k}^2}^{\frac 1 2}\) | Definition of $d_2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\sum_{k \mathop \ge 0} \paren {y_k - x_k}^2}^{\frac 1 2}\) | Definition of Absolute Value | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map {d_2} {y, x}\) | Definition of $d_2$ |
So Metric Space Axiom $(\text M 3)$ holds for $d_2$.
$\Box$
Proof of Metric Space Axiom $(\text M 4)$
| \(\ds x\) | \(\ne\) | \(\ds y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists k \in \N: \, \) | \(\ds x_k\) | \(\ne\) | \(\ds y_k\) | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {x_k - y_k}^2\) | \(>\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {\sum_{k \mathop \ge 0} \paren {x_k - y_k}^2}^{\frac 1 2}\) | \(>\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map {d_2} {x, y}\) | \(>\) | \(\ds 0\) | Definition of $d_2$ |
So Metric Space Axiom $(\text M 4)$ holds for $d_2$.
$\blacksquare$