description/proof of that for separable Hilbert space and orthonormal subset, subset can be expanded to be orthonormal Schauder basis
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of Hilbert space.
- The reader knows a definition of topology induced by metric.
- The reader knows a definition of separable topological space.
- The reader knows a definition of Schauder Basis for Normed Vectors Space.
- The reader admits the proposition that for any Hilbert space, any countable orthonormal subset, and any element of the Hilbert space, the linear combination of the subset with the the-element-and-subset-element-inner-product coefficients converges.
- The reader admits the proposition that for any vectors space with the topology induced by the metric induced by the norm induced by any inner product, if the space is separable, it has no uncountable orthonormal subset.
- The reader admits the proposition that for any real or complex vectors space with the topology induced by the metric induced by the norm induced by any inner product, the inner product with any 1 argument fixed is a continuous map.
- The reader admits the proposition that for any continuous map and any net with directed index set that converges to any point on the domain, the image of the net converges to the image of the point and if the codomain is Hausdorff, the convergence of the image of the net is the image of the point.
- The reader admits the proposition that any separable Hilbert space has an orthonormal Schauder basis.
Target Context
- The reader will have a description and a proof of the proposition that for any separable Hilbert space and any orthonormal subset, the subset can be expanded to be an orthonormal Schauder basis.
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
\((V, dist)\): \(= \{\text{ the separable Hilbert spaces }\}\), with any inner product, \(\langle \bullet, \bullet \rangle\), with the topology induced by \(dist\)
\(S\): \(\in \{\text{ the orthonormal subsets of } V\}\)
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Statements:
\(\exists B \in \{\text{ the orthonormal Schauder bases for } V\} (S \subseteq B)\)
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2: Proof
Whole Strategy: Step 1: take the set of the orthonormal subsets that contain \(S\), \(S'\); Step 2: see that \(S'\) has a maximal element by Zorn's lemma; Step 3: see that the maximal element is an orthonormal Schauder basis for \(V\).
Step 1:
Any orthonormal subset, \(W \subseteq V\), means that for each \(w \in W\), \(\langle w, w \rangle = 1\) and for each \(w, w' \in W\) such that \(w \neq w'\), \(\langle w, w' \rangle = 0\).
Let us define the set of the orthonormal subsets that contain \(S\), \(S' := \{W \subseteq V \vert S \subseteq W \land W \in \{\text{ the orthonormal subsets of } V\}\}\), which is a well-defined set by the ZFC set theory, because the qualification, \(W \in \{\text{ the orthonormal subsets of } V\}\) is expressed by the legitimate formula, \(\forall w \in W (\langle w, w \rangle = 1 \land (\forall w' \in W \setminus \{w\} (\langle w, w' \rangle = 0)))\).
Step 2:
\(S' \neq \emptyset\), because \(S \in S'\).
\(S'\) satisfies the conditions for Zorn's lemma: for each chain, \(C \subseteq S'\), \(\cup C \in S'\), because while \(C\)' s being a chain means that for each \(c, c' \in C\), \(c \subseteq c'\) or \(c' \subseteq c\), for each \(p \in \cup C\), \(p \in c\) for a \(c \in C\), but \(c \in S'\), so, \(\langle p, p \rangle = 1\); for each \(p, p' \in \cup C\) such that \(p \neq p'\), \(p \in c\) for a \(c \in C\) and \(p' \in c'\) for a \(c' \in C\), but \(c \subseteq c'\) without loss of generality, so, \(p, p' \in c'\), but as \(c' \in S'\), \(\langle p, p' \rangle = 0\), while \(S \subseteq \cup C\).
So, by Zorn's lemma, \(S'\) has a maximal element, \(B \in S'\).
Step 3:
Let us see that \(B\) is an orthonormal Schauder basis for \(V\).
\(B\) is orthonormal.
\(B\) is countable, by the proposition that for any vectors space with the topology induced by the metric induced by the norm induced by any inner product, if the space is separable, it has no uncountable orthonormal subset, so, let us denote \(B = \{b_1, b_2, ...\}\).
For each \(v \in V\), \(\sum_j \langle v, b_j \rangle b_j\) converges, by the proposition that for any Hilbert space, any countable orthonormal subset, and any element of the Hilbert space, the linear combination of the subset with the the-element-and-subset-element-inner-product coefficients converges.
\(v - \sum_j \langle v, b_j \rangle b_j\) is orthogonal to \(B\), because for each \(b_l \in B\), \(\langle v - \sum_j \langle v, b_j \rangle b_j, b_l \rangle = \langle v - lim_n \sum^n_{j = 1} \langle v, b_j \rangle b_j, b_l \rangle = \langle lim_n (v - \sum^n_{j = 1} \langle v, b_j \rangle b_j), b_l \rangle = lim_n \langle v - \sum^n_{j = 1} \langle v, b_j \rangle b_j, b_l \rangle\), by the proposition that for any real or complex vectors space with the topology induced by the metric induced by the norm induced by any inner product, the inner product with any 1 argument fixed is a continuous map and the proposition that for any continuous map and any net with directed index set that converges to any point on the domain, the image of the net converges to the image of the point and if the codomain is Hausdorff, the convergence of the image of the net is the image of the point, \(= lim_n (\langle v, b_l \rangle - \langle v, b_l \rangle) = lim_n 0 = 0\).
Note that for each \(v' \in V\) that is orthogonal to \(B\), \(v' = 0\), because otherwise, \(v'\) could be normalized to be \(b'\) and \(B \cup \{b'\}\) would be an element larger than \(B\) in \(S'\), a contradiction against \(B\)' s being maximal.
So, \(v - \sum_j \langle v, b_j \rangle b_j = 0\), which means that \(v = \sum_j \langle v, b_j \rangle b_j\).
So, each \(v \in V\) is expressed as \(v = \sum_j \langle v, b_j \rangle b_j\).
Let us see that the decomposition is unique (this is the same logic done in the proposition that any separable Hilbert space has an orthonormal Schauder basis).
1st, let us see that for each \(v = \sum_j v^j b_j\) and \(v' = \sum_j v'^j b_j\), \(v' - v = \sum_j (v'^j - v^j) b_j\).
\(\Vert v' - v - \sum^n_{j = 1} (v'^j - v^j) b_j \Vert = \Vert (v' - \sum^n_{j = 1} v'^j b_j) - (v - \sum^n_{j = 1} v^j b_j) \Vert \le \Vert v' - \sum^n_{j = 1} v'^j b_j \Vert + \Vert v - \sum^n_{j = 1} v^j b_j \Vert\), but for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is an \(N' \subseteq \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N' \lt n\), \(\Vert v' - \sum^n_{j = 1} v'^j b_j \Vert \lt \epsilon / 2\) and there is an \(N \subseteq \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N \lt n\), \(\Vert v - \sum^n_{j = 1} v^j b_j \Vert \lt \epsilon / 2\), and so, we can take \(max (N', N)\) and for each \(n \in \mathbb{N}\) such that \(max (N', N) \lt n\), \(\Vert v' - \sum^n_{j = 1} v'^j b_j \Vert + \Vert v - \sum^n_{j = 1} v^j b_j \Vert \lt \epsilon / 2 + \epsilon / 2 = \epsilon\), so, \(\Vert v' - v - \sum^n_{j = 1} (v'^j - v^j) b_j \Vert \lt \epsilon\), which means that \(v' - v = \sum_j (v'^j - v^j) b_j\).
2nd, let us see that \(v = \sum_j v^j b_j = 0\) implies that \(v^j = 0\).
If \(v^l \neq 0\), \(\langle v, b_l \rangle = \langle \sum_j v^j b_j, b_l \rangle = \langle lim_n \sum^n_{j = 1} v^j b_j, b_l \rangle = lim_n \langle \sum^n_{j = 1} v^j b_j, b_l \rangle = lim_n \sum^n_{j = 1} v^j \langle b_j, b_l \rangle = lim_n \sum^n_{j = 1} v^j \delta_{j, l} = lim_n v^l \neq 0\), a contradiction against \(v = 0\): for \(\langle 0, b_l \rangle = c\), \(\langle 0, b_l \rangle = \langle 0 + 0, b_l \rangle = \langle 0, b_l \rangle + \langle 0, b_l \rangle = c + c = 2 c = c\), so, \(c = 0\).
Now, if there is a \(v = \sum_j v^j b_j\), \(0 = v - v = \sum_j (v^j - \langle v, b_j \rangle) b_j\), which implies that \(v^j = \langle v, b_j \rangle\).
So, \(B\) is an orthonormal Schauder basis.
3: Note
A typical case to which this proposition can be applied is that there is a Hilbert subspace of \(V\) (any closed subspace of \(V\) is one by the proposition that for any complete metric space, any closed subspace is complete), which is inevitably a separable Hilbert space by the proposition that for any separable topological space induced by any metric, any topological subspace is separable, so, an orthonormal Schauder basis for the subspace can be taken by the proposition that any separable Hilbert space has an orthonormal Schauder basis, and it can be expanded to be an orthonormal Schauder basis for \(V\), by this proposition.
