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

2. The attempt at a solution
If A is symmetric, then A\subseteq A^*, where A^* is the adjoint of A, and from A\supseteq B one can deduce that also B is symmetric. The definition domains of A and B are dense in \mathcal{H} (\mathcal{H} an Hilbert space), so \mathcal{N}_{(A+\imath I)^*}=(\mathcal{R}_{A+\imath I})^\perp, where now \mathcal{N} is the operator's kernel. An idea to complete the exercise should be showing that A^*\supseteq B^*, using the identity \mathcal{N}_{(A+\imath I)^*}=\mathcal{N}_{(B+\imath I)^*} 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 \mathcal{R}_{A+iI}=\mathcal{R}_{B+iI} tell you now?

 

[johan_munchen]

For hypothesis A\supseteq B so \mathcal{D}_A\supseteq\mathcal{D}_B and A\mathbf{x}=B\mathbf{x}\,\,\forall\,\mathbf{x}\in \mathcal{D}_B. Let \mathbf{a}\in\mathcal{D}_A then A\mathbf{a}+\imath \mathbf{a}\in\mathcal{H}. For hypothesis \mathcal{R}_{A+\imath I}=\mathcal{R}_{B+\imath I}, so \exists\,\mathbf{y}\in\mathcal{H}\,:\,\mathbf{y}=A\mathbf{a}+\imath I\mathbf{a}=B\mathbf{a}+\imath I\mathbf{a}, then B\mathbf{a}+\imath I\mathbf{a}\in\mathcal{H}, which means that \mathbf{a} is also an element of \mathcal{D}_B. So \mathcal{D}_A=\mathcal{D}_B, in other words A=B. I have thought about this but I was trying to use the symmetry hypotesis about A.