Proving Symmetric Operators are Equal: A Functional Analysis Challenge

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Homework Help Overview

The problem involves proving that two symmetric operators, A and B, are equal under certain conditions related to their ranges and domains in the context of functional analysis.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of the symmetry of operators and the relationship between their ranges. There is an exploration of the definitions and properties of the operators involved, particularly focusing on the adjoint and kernel. Some participants question how certain identities and hypotheses can be utilized effectively in the proof.

Discussion Status

The discussion is ongoing, with various participants offering insights and attempting to clarify the relationships between the operators. Some guidance has been provided regarding the implications of the range equality, but there is no explicit consensus on the approach to take.

Contextual Notes

There are references to the dense domains of the operators and the specific properties of symmetric operators, which may influence the discussion. Participants also note the distinction between functional analysis and introductory physics, indicating a potential misunderstanding of the subject area.

johan_munchen
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A functional analysis' problem

I hope this is the right place to submit this post.

Homework Statement


Let [itex]A[/itex] be a symmetric operator, [itex]A\supseteq B[/itex] and [itex]\mathcal{R}_{A+\imath I}=\mathcal{R}_{B+\imath I}[/itex] (where [itex]\mathcal{R}[/itex] means the range of the operator). Show that [itex]A=B[/itex].

2. The attempt at a solution
If [itex]A[/itex] is symmetric, then [itex]A\subseteq A^*[/itex], where [itex]A^*[/itex] is the adjoint of [itex]A[/itex], and from [itex]A\supseteq B[/itex] one can deduce that also [itex]B[/itex] is symmetric. The definition domains of [itex]A[/itex] and [itex]B[/itex] are dense in [itex]\mathcal{H}[/itex] ([itex]\mathcal{H}[/itex] an Hilbert space), so [itex]\mathcal{N}_{(A+\imath I)^*}=(\mathcal{R}_{A+\imath I})^\perp[/itex], where now [itex]\mathcal{N}[/itex] is the operator's kernel. An idea to complete the exercise should be showing that [itex]A^*\supseteq B^*[/itex], using the identity [itex]\mathcal{N}_{(A+\imath I)^*}=\mathcal{N}_{(B+\imath I)^*}[/itex] and the previous hypothesis. However, I can't understand how this could be usefull.

Thanks a lot for your help. JM
 
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Is it possible to move this post to the "Introductory Physics" section?
 


johan_munchen said:
Is it possible to move this post to the "Introductory Physics" section?

No need, functional analysis is mathematics. :smile:

Anyway, take a point a in the domain of definition of A. Let's calculate Aa+ia, what does [itex]\mathcal{R}_{A+iI}=\mathcal{R}_{B+iI}[/itex] tell you now?
 
For hypothesis [itex]A\supseteq B[/itex] so [itex]\mathcal{D}_A\supseteq\mathcal{D}_B[/itex] and [itex]A\mathbf{x}=B\mathbf{x}\,\,\forall\,\mathbf{x}\in \mathcal{D}_B[/itex]. Let [itex]\mathbf{a}\in\mathcal{D}_A[/itex] then [itex]A\mathbf{a}+\imath \mathbf{a}\in\mathcal{H}[/itex]. For hypothesis [itex]\mathcal{R}_{A+\imath I}=\mathcal{R}_{B+\imath I}[/itex], so [itex]\exists\,\mathbf{y}\in\mathcal{H}\,:\,\mathbf{y}=A\mathbf{a}+\imath I\mathbf{a}=B\mathbf{a}+\imath I\mathbf{a}[/itex], then [itex]B\mathbf{a}+\imath I\mathbf{a}\in\mathcal{H}[/itex], which means that [itex]\mathbf{a}[/itex] is also an element of [itex]\mathcal{D}_B[/itex]. So [itex]\mathcal{D}_A=\mathcal{D}_B[/itex], in other words [itex]A=B[/itex]. I have thought about this but I was trying to use the symmetry hypotesis about A.
 
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