definition of finite-product Hilbert space
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of Hilbert space.
- The reader knows a definition of finite-product inner product.
Target Context
- The reader will have a definition of finite-product Hilbert space.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\( \{V_1, ..., V_n\}\): \(\subseteq \{\text{ the } F \text{ Hilbert spaces }\}\), with any inner products, \(\{\langle \bullet, \bullet \rangle_1, ..., \langle \bullet, \bullet \rangle_n\}\)
\(*V_1 \times ... \times V_n\): \(= \text{ the product vectors space with the product inner product }\) with the metric induced by the norm induced by the inner product, \(\in \{\text{ the } F \text{ Hilbert spaces }\}\)
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Conditions:
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2: Note
Let us see that \(V_1 \times ... \times V_n\) is indeed a complete metric space, so, indeed a Hilbert space.
Let \(s: \mathbb{N} \to V_1 \times ... \times V_n\) be any Cauchy sequence.
For each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is an \(N \in \mathbb{N}\) such that for each \(m, o \in \mathbb{N}\) such that \(N \lt m, o\), \(\langle s (o) - s (m), s (o) - s (m) \rangle \lt \epsilon^2\).
\(\langle s (o) - s (m), s (o) - s (m) \rangle = \langle (s (o)^1, ..., s (o)^n) - (s (m)^1, ..., s (m)^n), (s (o)^1, ..., s (o)^n) - (s (m)^1, ..., s (m)^n) \rangle = \langle (s (o)^1 - s (m)^1, ..., s (o)^n - s (m)^n), (s (o)^1 - s (m)^1, ..., s (o)^n - s (m)^n) \rangle = \langle s (o)^1 - s (m)^1, s (o)^1 - s (m)^1 \rangle_1 + ... + \langle s (o)^n - s (m)^n, s (o)^n - s (m)^n \rangle_n \lt \epsilon^2\), which means that for each \(j \in \{1, ..., n\}\), \(\langle s (o)^j - s (m)^j, s (o)^j - s (m)^j \rangle_j \lt \epsilon^2\).
That means that \(s^j: \mathbb{N} \to V_j\) is a Cauchy sequence.
As \(V_j\) is complete, \(s^j\) converges to a \(v^j \in V_j\).
Let \(v := (v^1, ..., v^n) \in V_1 \times ... \times V_n\).
\(s\) converges to \(v\), because for each \(\epsilon\), we can take an \(N\) such that for each \(m\) such that \(N \lt m\) and each \(j\), \(\langle v^j - s (m)^j \rangle \lt \epsilon^2 / n\), and then, \(\langle v - s (m), v - s (m) \rangle = \langle (v^1, ..., v^n) - (s (m)^1, ..., s (m)^n), (v^1, ..., v^n) - (s (m)^1, ..., s (m)^n) \rangle = \langle v^1 - s (m)^1, v^1 - s (m)^1 \rangle_1 + ... \langle v^n - s (m)^n, v^n - s (m)^n \rangle_n \lt \epsilon^2 / n + ... + \epsilon^2 / n = \epsilon^2\).
