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!


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}

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.