1095: For Hilbert Space, Countable Orthonormal Subset, and Element of Hilbert Space, Linear Combination of Subset with Element-And-Subset-Element-Inner-Product Coefficients Converges

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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

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of Hilbert space.

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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.

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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.

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Main Body

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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, $⟨\bullet ,\bullet ⟩$
$S$: $\in \{\text{ the countable subsets of }V\}$, $=\{s_1,s_2,...\}$, such that $\mathrm{\forall }{s}_j\in S(⟨s_j,s_j⟩=1)\wedge \mathrm{\forall }{s}_j,s_l\in S\text{ such that }{s}_j\ne s_l(⟨s_j,s_l⟩=0)$
$v$: $\in V$
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Statements:
$\sum _j⟨v,s_j⟩s_j$ converges
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2: Proof

Whole Strategy: Step 1: see that $\sum _j⟨v,s_j⟩s_j$ is a Cauchy sequence; Step 2: conclude the proposition.

Step 1:

Let us see that $\sum _j⟨v,s_j⟩s_j$ is a Cauchy sequence.

$0\le \Vert v-\sum _{j=1}^n⟨v,s_j⟩s_j{\Vert }^2=⟨v-\sum _{j=1}^n⟨v,s_j⟩s_j,v-\sum _{l=1}^n⟨v,s_l⟩s_l⟩=⟨v,v⟩-⟨v,\sum _{l=1}^n⟨v,s_l⟩s_l⟩-⟨\sum _{j=1}^n⟨v,s_j⟩s_j,v⟩+⟨\sum _{j=1}^n⟨v,s_j⟩s_j,\sum _{l=1}^n⟨v,s_l⟩s_l⟩=⟨v,v⟩-\sum _{l=1}^n\overline{⟨v,s_l⟩}⟨v,s_l⟩-\sum _{j=1}^n⟨v,s_j⟩⟨s_j,v⟩+\sum _{j=1}^n⟨v,s_j⟩\sum _{l=1}^n\overline{⟨v,s_l⟩}⟨s_j,s_l⟩=⟨v,v⟩-\sum _{l=1}^n|⟨v,s_l⟩{|}^2-\sum _{j=1}^n|⟨v,s_j⟩{|}^2+\sum _{j=1}^n⟨v,s_j⟩\sum _{l=1}^n\overline{⟨v,s_l⟩}{\delta }_{j,l}=⟨v,v⟩-\sum _{j=1}^n|⟨v,s_j⟩{|}^2-\sum _{j=1}^n|⟨v,s_j⟩{|}^2+\sum _{j=1}^n⟨v,s_j⟩\overline{⟨v,s_j⟩}=⟨v,v⟩-2\sum _{j=1}^n|⟨v,s_j⟩{|}^2+\sum _{j=1}^n|⟨v,s_j⟩{|}^2=⟨v,v⟩-\sum _{j=1}^n|⟨v,s_j⟩{|}^2$, which means that $\sum _{j=1}^n|⟨v,s_j⟩{|}^2\le ⟨v,v⟩$.

That means that $\sum _{j=1}^n|⟨v,s_j⟩{|}^2$ converges as a monotone-increasing upper-bounded sequence, and so, is a Cauchy sequence, which means that for each $ϵ$, there is an $N$ such that for each $N<m,n$, $\sum _{j=m}^n|⟨v,s_j⟩{|}^2<{ϵ}^2$.

$\Vert \sum _{j=m}^n⟨v,s_j⟩s_j{\Vert }^2=⟨\sum _{j=m}^n⟨v,s_j⟩s_j,\sum _{l=m}^n⟨v,s_l⟩s_l⟩=\sum _{j=m}^n⟨v,s_j⟩\sum _{l=m}^n\overline{⟨v,s_l⟩}⟨s_j,s_l⟩=\sum _{j=m}^n⟨v,s_j⟩\sum _{l=m}^n\overline{⟨v,s_l⟩}{\delta }_{j,l}=\sum _{j=m}^n|⟨v,s_j⟩{|}^2<{ϵ}^2$.

That means that $\sum _{j=1}^n⟨v,s_j⟩s_j$ is a Cauchy sequence.

Step 2:

As $V$ is complete, $\sum _j⟨v,s_j⟩s_j$ converges.

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References

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