description/proof of that for Hilbert space, countable orthonormal subset, and element of Hilbert space, linear combination of subset with element-and-subset-element-inner-product coefficients converges
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of Hilbert space.
Target Context
- The reader will have a description and a proof of 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.
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
\((V, dist)\): \(= \{\text{ the Hilbert spaces }\}\), with any inner product, \(\langle \bullet, \bullet \rangle\)
\(S\): \(\in \{\text{ the countable subsets of } V\}\), \(= \{s_1, s_2, ...\}\), such that \(\forall s_j \in S (\langle s_j, s_j \rangle = 1) \land \forall s_j, s_l \in S \text{ such that } s_j \neq s_l (\langle s_j, s_l \rangle = 0)\)
\(v\): \(\in V\)
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Statements:
\(\sum_j \langle v, s_j \rangle s_j\) converges
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2: Proof
Whole Strategy: Step 1: see that \(\sum_j \langle v, s_j \rangle s_j\) is a Cauchy sequence; Step 2: conclude the proposition.
Step 1:
Let us see that \(\sum_j \langle v, s_j \rangle s_j\) is a Cauchy sequence.
\(0 \le \Vert v - \sum^n_{j = 1} \langle v, s_j \rangle s_j \Vert^2 = \langle v - \sum^n_{j = 1} \langle v, s_j \rangle s_j, v - \sum^n_{l = 1} \langle v, s_l \rangle s_l \rangle = \langle v, v \rangle - \langle v, \sum^n_{l = 1} \langle v, s_l \rangle s_l\rangle - \langle \sum^n_{j = 1} \langle v, s_j \rangle s_j, v \rangle + \langle \sum^n_{j = 1} \langle v, s_j \rangle s_j, \sum^n_{l = 1} \langle v, s_l \rangle s_l \rangle = \langle v, v \rangle - \sum^n_{l = 1} \overline{\langle v, s_l \rangle} \langle v, s_l \rangle - \sum^n_{j = 1} \langle v, s_j \rangle \langle s_j, v \rangle + \sum^n_{j = 1} \langle v, s_j \rangle \sum^n_{l = 1} \overline{\langle v, s_l \rangle} \langle s_j, s_l \rangle = \langle v, v \rangle - \sum^n_{l = 1} \vert \langle v, s_l\rangle \vert^2 - \sum^n_{j = 1} \vert \langle v, s_j \rangle \vert^2 + \sum^n_{j = 1} \langle v, s_j \rangle \sum^n_{l = 1} \overline{\langle v, s_l \rangle} \delta_{j, l} = \langle v, v \rangle - \sum^n_{j = 1} \vert \langle v, s_j \rangle \vert^2 - \sum^n_{j = 1} \vert \langle v, s_j \rangle \vert^2 + \sum^n_{j = 1} \langle v, s_j \rangle \overline{\langle v, s_j \rangle} = \langle v, v \rangle - 2 \sum^n_{j = 1} \vert \langle v, s_j \rangle \vert^2 + \sum^n_{j = 1} \vert \langle v, s_j \rangle \vert^2 = \langle v, v \rangle - \sum^n_{j = 1} \vert \langle v, s_j \rangle \vert^2\), which means that \(\sum^n_{j = 1} \vert \langle v, s_j \rangle \vert^2 \le \langle v, v \rangle\).
That means that \(\sum^n_{j = 1} \vert \langle v, s_j \rangle \vert^2\) converges as a monotone-increasing upper-bounded sequence, and so, is a Cauchy sequence, which means that for each \(\epsilon\), there is an \(N\) such that for each \(N \lt m, n\), \(\sum^n_{j = m} \vert \langle v, s_j \rangle \vert^2 \lt \epsilon^2\).
\(\Vert \sum^n_{j = m} \langle v, s_j \rangle s_j \Vert^2 = \langle \sum^n_{j = m} \langle v, s_j \rangle s_j, \sum^n_{l = m} \langle v, s_l \rangle s_l \rangle = \sum^n_{j = m} \langle v, s_j \rangle \sum^n_{l = m} \overline{\langle v, s_l \rangle} \langle s_j, s_l \rangle = \sum^n_{j = m} \langle v, s_j \rangle \sum^n_{l = m} \overline{\langle v, s_l \rangle} \delta_{j, l} = \sum^n_{j = m} \vert \langle v, s_j \rangle \vert^2 \lt \epsilon^2\).
That means that \(\sum^n_{j = 1} \langle v, s_j \rangle s_j\) is a Cauchy sequence.
Step 2:
As \(V\) is complete, \(\sum_j \langle v, s_j \rangle s_j\) converges.
