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I am trying to answer the following question:

Let $A, B$ be two linear, closed, densely defined operators in a Hilbert space $H$ such that $D(A)=D(B)=D$ and $(Ax,y)=(x,By)$ for every $x,y\in D$. Can we infer that $B=A^*$?

To show $A^*=B$, I only need to prove that $D(A^*)\subset D$. But I can neither prove this or find a counterexample.

Help me, please. Many thanks!

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Let $A$ be a closed densely-defined symmetric linear operator on a Hilbert space $H$ that is not selfadjoint. Let $B=A$. Then $(Ax,y)=(x,By)$ for all $x,y\in\mathcal{D}(A)=\mathcal{D}(B)$ because $A$ is symmetric. However, $A^{\star} \ne B$ because $A^*\ne A$, as $A$ is not selfadjoint.

Example: Let $H=L^2[0,1]$. Let $A=\frac{1}{i}\frac{d}{dx}$ on the domain consisting of $f \in L^2[0,1]$ that is equal a.e. to an absolutely continuous function $f_a$ on $[0,1]$ for which $f_a'\in L^2$, and such that $f_a(0)=f_a(1)=0$. Let $B=A$. Then $A$ is closed and densely-defined, with $$ (Af,g)-(f,Bg) = \frac{1}{i}\int_{0}^{1}f'\overline{g}+f\overline{g}'dt=0,\;\;\; f,g\in\mathcal{D}(A)=\mathcal{D}(B). $$ However, $B \ne A^*$ because the constant function $1$ is in the domain of $A^*$ but not in the domain of $B$, as seen from \begin{align} (Af,1) & = \frac{1}{i}\int_{0}^{1}f'dt = \frac{1}{i}[f_a(1)-f_a(0)]=0\\ & \implies (Af,1)=(f,0),\;\; f\in\mathcal{D}(A) \\ & \implies 1\in\mathcal{D}(A^*) \mbox{ and } A^*1 = 0. \end{align}

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  • $\begingroup$ Thank you very much. I understand now. I want to vote for your answer but I don't know how to do. Can you show me? $\endgroup$ – T. M. Nguyet Oct 27 '16 at 9:38
  • $\begingroup$ @T.M.Nguyet : There is a checkmark to the left of the question that you can select and turn it green. $\endgroup$ – Disintegrating By Parts Oct 27 '16 at 14:00
  • $\begingroup$ Maybe he wants a concrete example. Admittedly, most people new to the field don't know the difference between a symmetric operator and a self adjoint one. Though, I did enjoy your answer. $\endgroup$ – Squirtle Oct 27 '16 at 19:36
  • $\begingroup$ @Squirtle : Good idea. I added that. $\endgroup$ – Disintegrating By Parts Oct 27 '16 at 20:32
  • $\begingroup$ @ TrialAndError: Your example is very helpful and easy to understand. Thanks. Could you show me if you have an example in the case that H is a real Hilbert space? Thank again! $\endgroup$ – T. M. Nguyet Oct 28 '16 at 2:52
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Let $f(x) = (Ax,y)$ and let $g(x) = (x,By)$, then clearly $f$ and $g$ are continuous. We assumed that $f(x)=g(x)$ for all $x\in \mathcal{D}$ and by the following

$f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$

we conclude $f=g$ for all $x\in \mathcal{H}$ so we are finished.

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  • $\begingroup$ Thank you for your answer. But $\endgroup$ – T. M. Nguyet Oct 23 '16 at 2:53
  • $\begingroup$ Thank you for your help. But I really don't understand why $f,g$ are continuous on $H$. They are only defined on $D$, so we can't use closed graph theorem. Could you explain this for me? Thanks again! $\endgroup$ – T. M. Nguyet Oct 23 '16 at 3:03
  • $\begingroup$ $f$ is just an inner product with $A$ and $y$ fixed. It is a standard fact that this is continuous and many books on functional analysis have the proof early on. You can also check this out: math.stackexchange.com/questions/4501/… Hopefully, this answers your question; if it does.... please vote; if it doesn't please just ask. $\endgroup$ – Squirtle Oct 23 '16 at 3:52
  • $\begingroup$ I think that argument doesn't work in this situation, because we need $A$ is bounded and defined on the whole space to prove the continuity of $f,g$, if we follow that way. $\endgroup$ – T. M. Nguyet Oct 23 '16 at 8:40
  • $\begingroup$ Oh! Yes. Somehow I confused closed and continuous. I am going to leave my answer up (in case it helps anyone) for the more special case I assumed tacitly. I will try to solve the more general problem today. $\endgroup$ – Squirtle Oct 23 '16 at 16:32

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