Proving Symmetric Operators are Equal: A Functional Analysis Challenge

[johan_munchen]

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

 

[johan_munchen]

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

 

[micromass]

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

No need, functional analysis is mathematics.

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?

 

[johan_munchen]

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.