Hilbert Sequence Space is Metric Space

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $A$ be the set of all real sequences $\left\langle{x_i}\right\rangle$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent.

Let $\ell^2 = \left({A, d_2}\right)$ 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 $\left\langle{x_i}\right\rangle$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent.

Then $\ell^2 := \left({A, d_2}\right)$ where $d_2: A \times A: \to \R$ is the real-valued function defined as:

$\displaystyle \forall x = \left\langle{x_i}\right\rangle, y = \left\langle{y_i}\right\rangle \in A: d_2 \left({x, y}\right) := \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2}$


From Convergence of Square of Linear Combination of Sequences whose Squares Converge we have that $\displaystyle \sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2$ does actually converge.


Proof of $M1$

\(\displaystyle d_2 \left({x, x}\right)\) \(=\) \(\displaystyle \left({\sum_{k \mathop \ge 0} \left({x_k - x_k}\right)^2}\right)^{\frac 1 2}\) Definition of $d_2$
\(\displaystyle \) \(=\) \(\displaystyle \left({\sum_{k \mathop \ge 0} 0^2}\right)^{\frac 1 2}\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)

So axiom $M1$ holds for $d_2$.

$\Box$


Proof of $M2$

Let $z = \left\langle{z_i}\right\rangle \in A$.


\(\displaystyle d_2 \left({x, y}\right) + d_2 \left({y, z}\right)\) \(=\) \(\displaystyle \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2} + \left({\sum_{k \mathop \ge 0} \left({y_k - z_k}\right)^2}\right)^{\frac 1 2}\) Definition of $d_2$
\(\displaystyle \) \(\ge\) \(\displaystyle \left({\sum_{k \mathop \ge 0} \left({x_k - z_k}\right)^2}\right)^{\frac 1 2}\) Minkowski's Inequality for Sums: index $2$
\(\displaystyle \) \(=\) \(\displaystyle d_2 \left({x, z}\right)\) Definition of $d_2$

So axiom $M2$ holds for $d_2$.

$\Box$


Proof of $M3$

\(\displaystyle d_2 \left({x, y}\right)\) \(=\) \(\displaystyle \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2}\) Definition of $d_2$
\(\displaystyle \) \(=\) \(\displaystyle \left({\sum_{k \mathop \ge 0} \left({y_k - x_k}\right)^2}\right)^{\frac 1 2}\) Definition of Absolute Value
\(\displaystyle \) \(=\) \(\displaystyle d_2 \left({y, x}\right)\) Definition of $d_2$

So axiom $M3$ holds for $d_2$.

$\Box$


Proof of $M4$

\(\displaystyle x\) \(\ne\) \(\displaystyle y\)
\(\displaystyle \implies \ \ \) \(\, \displaystyle \exists k \in \N: \, \) \(\displaystyle x_k\) \(\ne\) \(\displaystyle y_k\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left({x_k - y_k}\right)^2\) \(>\) \(\displaystyle 0\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2}\) \(>\) \(\displaystyle 0\)
\(\displaystyle \implies \ \ \) \(\displaystyle d_2 \left({x, y}\right)\) \(>\) \(\displaystyle 0\) Definition of $d_2$

So axiom $M4$ holds for $d_2$.

$\blacksquare$