# A functional analysis problem

1. Jul 14, 2011

### johan_munchen

A functional analysis' problem

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

1. The problem statement, all variables and given/known data
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

Last edited: Jul 14, 2011
2. Jul 14, 2011

### johan_munchen

Re: A functional analysis' problem

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

3. Jul 14, 2011

### micromass

Re: A functional analysis' problem

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?

4. Jul 15, 2011

### 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 simmetry hypotesis about A.

Last edited: Jul 15, 2011