A multiplicative operator is self-adjoint












1












$begingroup$


I am doing the following problem: Let $(M,mathcal{A},mu)$ be a general measure space, $f:Mto mathbb{R}$ be a measurable function. Define the operator $A_f:mathrm{dom}(A_f)to L^2(mu)$ by $$mathrm{dom}(A_f):={psiin L^2(mu)|fpsiin L^2(mu)}$$ $$A_fpsi=fpsi,forallpsiinmathrm{dom}(A_f)$$ Show this operator is self-adjoint for any given measurable function $f$.



It is clear that this operator is symmetric, but I can't think of any method to argue the domain of $A^*_f$ is the same as that of $A_f$ (I know it's at least larger so we only need to prove $mathrm{dom}(A^*_f)subsetmathrm{dom}(A_f)$). By definition I can have $$left<A_f^*psi,phiright>=left<psi,A_fphiright>=left<psi,fphiright>=
left<fpsi,phiright>,forall phiinmathrm{dom}(A_f)$$
so $A^*_fpsi=fpsi,forall psiinmathrm{dom}(A^*_f)$.
However it doesn't seem to help much.






functional-analysis self-adjoint-operators







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    1












    $begingroup$


    I am doing the following problem: Let $(M,mathcal{A},mu)$ be a general measure space, $f:Mto mathbb{R}$ be a measurable function. Define the operator $A_f:mathrm{dom}(A_f)to L^2(mu)$ by $$mathrm{dom}(A_f):={psiin L^2(mu)|fpsiin L^2(mu)}$$ $$A_fpsi=fpsi,forallpsiinmathrm{dom}(A_f)$$ Show this operator is self-adjoint for any given measurable function $f$.



    It is clear that this operator is symmetric, but I can't think of any method to argue the domain of $A^*_f$ is the same as that of $A_f$ (I know it's at least larger so we only need to prove $mathrm{dom}(A^*_f)subsetmathrm{dom}(A_f)$). By definition I can have $$left<A_f^*psi,phiright>=left<psi,A_fphiright>=left<psi,fphiright>=
    left<fpsi,phiright>,forall phiinmathrm{dom}(A_f)$$
    so $A^*_fpsi=fpsi,forall psiinmathrm{dom}(A^*_f)$.
    However it doesn't seem to help much.






    functional-analysis self-adjoint-operators







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am doing the following problem: Let $(M,mathcal{A},mu)$ be a general measure space, $f:Mto mathbb{R}$ be a measurable function. Define the operator $A_f:mathrm{dom}(A_f)to L^2(mu)$ by $$mathrm{dom}(A_f):={psiin L^2(mu)|fpsiin L^2(mu)}$$ $$A_fpsi=fpsi,forallpsiinmathrm{dom}(A_f)$$ Show this operator is self-adjoint for any given measurable function $f$.



      It is clear that this operator is symmetric, but I can't think of any method to argue the domain of $A^*_f$ is the same as that of $A_f$ (I know it's at least larger so we only need to prove $mathrm{dom}(A^*_f)subsetmathrm{dom}(A_f)$). By definition I can have $$left<A_f^*psi,phiright>=left<psi,A_fphiright>=left<psi,fphiright>=
      left<fpsi,phiright>,forall phiinmathrm{dom}(A_f)$$
      so $A^*_fpsi=fpsi,forall psiinmathrm{dom}(A^*_f)$.
      However it doesn't seem to help much.






      functional-analysis self-adjoint-operators







      share|cite|improve this question









      $endgroup$




      I am doing the following problem: Let $(M,mathcal{A},mu)$ be a general measure space, $f:Mto mathbb{R}$ be a measurable function. Define the operator $A_f:mathrm{dom}(A_f)to L^2(mu)$ by $$mathrm{dom}(A_f):={psiin L^2(mu)|fpsiin L^2(mu)}$$ $$A_fpsi=fpsi,forallpsiinmathrm{dom}(A_f)$$ Show this operator is self-adjoint for any given measurable function $f$.



      It is clear that this operator is symmetric, but I can't think of any method to argue the domain of $A^*_f$ is the same as that of $A_f$ (I know it's at least larger so we only need to prove $mathrm{dom}(A^*_f)subsetmathrm{dom}(A_f)$). By definition I can have $$left<A_f^*psi,phiright>=left<psi,A_fphiright>=left<psi,fphiright>=
      left<fpsi,phiright>,forall phiinmathrm{dom}(A_f)$$
      so $A^*_fpsi=fpsi,forall psiinmathrm{dom}(A^*_f)$.
      However it doesn't seem to help much.






      functional-analysis self-adjoint-operators




      functional-analysis self-adjoint-operators






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 7 '18 at 11:46









      ApocalypseApocalypse

      1378




      1378






















          1 Answer
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          1












          $begingroup$

          If you can show $A_f^ast psi=fpsi$, you are done (since $A_f^ast psiin L^2$ by definition). However, this does not follow immediately from the equality
          $$
          langle A_f^ast psi,phirangle=int fpsiphi,dmu.
          $$
          because $fpsi$ is a priori only a measurable function and not in $L^2$ (that's what you want to prove).



          I think the easiest way to prove self-adjointness here is to show that $A_fpm i$ is invertible. Just verify that $A_{(fpm i)^{-1}}$ is the inverse.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I am not so clear about that, nothing requires $A_f^*psi$ to be in $L^2$.
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:23










          • $begingroup$
            We just know $psi$ is in the domain of $A_f^*$, but it doesn't mean $A_f^*psi$ is also in $L^2$. Though the definition of $A_f$ requires $fphi$ to be square integrable given $phiinmathrm{dom}(A_f)$
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:27










          • $begingroup$
            The adjoint is an (unbounded) operator on $L^2$. In particular, it maps into $L^2$.
            $endgroup$
            – MaoWao
            Dec 7 '18 at 12:27












          • $begingroup$
            I mean even with $A^*_fpsi=fpsi$. It is still not valid to show $A_f^*psiin L^2$
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:28












          • $begingroup$
            Once again: If $T$ is a densely defined operator on the Hilbert space $H$, then $T^ast$ is an operator on $H$. This is just the definition of the adjoint.
            $endgroup$
            – MaoWao
            Dec 7 '18 at 12:30











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          $begingroup$

          If you can show $A_f^ast psi=fpsi$, you are done (since $A_f^ast psiin L^2$ by definition). However, this does not follow immediately from the equality
          $$
          langle A_f^ast psi,phirangle=int fpsiphi,dmu.
          $$
          because $fpsi$ is a priori only a measurable function and not in $L^2$ (that's what you want to prove).



          I think the easiest way to prove self-adjointness here is to show that $A_fpm i$ is invertible. Just verify that $A_{(fpm i)^{-1}}$ is the inverse.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I am not so clear about that, nothing requires $A_f^*psi$ to be in $L^2$.
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:23










          • $begingroup$
            We just know $psi$ is in the domain of $A_f^*$, but it doesn't mean $A_f^*psi$ is also in $L^2$. Though the definition of $A_f$ requires $fphi$ to be square integrable given $phiinmathrm{dom}(A_f)$
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:27










          • $begingroup$
            The adjoint is an (unbounded) operator on $L^2$. In particular, it maps into $L^2$.
            $endgroup$
            – MaoWao
            Dec 7 '18 at 12:27












          • $begingroup$
            I mean even with $A^*_fpsi=fpsi$. It is still not valid to show $A_f^*psiin L^2$
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:28












          • $begingroup$
            Once again: If $T$ is a densely defined operator on the Hilbert space $H$, then $T^ast$ is an operator on $H$. This is just the definition of the adjoint.
            $endgroup$
            – MaoWao
            Dec 7 '18 at 12:30
















          1












          $begingroup$

          If you can show $A_f^ast psi=fpsi$, you are done (since $A_f^ast psiin L^2$ by definition). However, this does not follow immediately from the equality
          $$
          langle A_f^ast psi,phirangle=int fpsiphi,dmu.
          $$
          because $fpsi$ is a priori only a measurable function and not in $L^2$ (that's what you want to prove).



          I think the easiest way to prove self-adjointness here is to show that $A_fpm i$ is invertible. Just verify that $A_{(fpm i)^{-1}}$ is the inverse.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I am not so clear about that, nothing requires $A_f^*psi$ to be in $L^2$.
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:23










          • $begingroup$
            We just know $psi$ is in the domain of $A_f^*$, but it doesn't mean $A_f^*psi$ is also in $L^2$. Though the definition of $A_f$ requires $fphi$ to be square integrable given $phiinmathrm{dom}(A_f)$
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:27










          • $begingroup$
            The adjoint is an (unbounded) operator on $L^2$. In particular, it maps into $L^2$.
            $endgroup$
            – MaoWao
            Dec 7 '18 at 12:27












          • $begingroup$
            I mean even with $A^*_fpsi=fpsi$. It is still not valid to show $A_f^*psiin L^2$
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:28












          • $begingroup$
            Once again: If $T$ is a densely defined operator on the Hilbert space $H$, then $T^ast$ is an operator on $H$. This is just the definition of the adjoint.
            $endgroup$
            – MaoWao
            Dec 7 '18 at 12:30














          1












          1








          1





          $begingroup$

          If you can show $A_f^ast psi=fpsi$, you are done (since $A_f^ast psiin L^2$ by definition). However, this does not follow immediately from the equality
          $$
          langle A_f^ast psi,phirangle=int fpsiphi,dmu.
          $$
          because $fpsi$ is a priori only a measurable function and not in $L^2$ (that's what you want to prove).



          I think the easiest way to prove self-adjointness here is to show that $A_fpm i$ is invertible. Just verify that $A_{(fpm i)^{-1}}$ is the inverse.






          share|cite|improve this answer









          $endgroup$



          If you can show $A_f^ast psi=fpsi$, you are done (since $A_f^ast psiin L^2$ by definition). However, this does not follow immediately from the equality
          $$
          langle A_f^ast psi,phirangle=int fpsiphi,dmu.
          $$
          because $fpsi$ is a priori only a measurable function and not in $L^2$ (that's what you want to prove).



          I think the easiest way to prove self-adjointness here is to show that $A_fpm i$ is invertible. Just verify that $A_{(fpm i)^{-1}}$ is the inverse.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 12:15









          MaoWaoMaoWao

          2,743617




          2,743617












          • $begingroup$
            I am not so clear about that, nothing requires $A_f^*psi$ to be in $L^2$.
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:23










          • $begingroup$
            We just know $psi$ is in the domain of $A_f^*$, but it doesn't mean $A_f^*psi$ is also in $L^2$. Though the definition of $A_f$ requires $fphi$ to be square integrable given $phiinmathrm{dom}(A_f)$
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:27










          • $begingroup$
            The adjoint is an (unbounded) operator on $L^2$. In particular, it maps into $L^2$.
            $endgroup$
            – MaoWao
            Dec 7 '18 at 12:27












          • $begingroup$
            I mean even with $A^*_fpsi=fpsi$. It is still not valid to show $A_f^*psiin L^2$
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:28












          • $begingroup$
            Once again: If $T$ is a densely defined operator on the Hilbert space $H$, then $T^ast$ is an operator on $H$. This is just the definition of the adjoint.
            $endgroup$
            – MaoWao
            Dec 7 '18 at 12:30


















          • $begingroup$
            I am not so clear about that, nothing requires $A_f^*psi$ to be in $L^2$.
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:23










          • $begingroup$
            We just know $psi$ is in the domain of $A_f^*$, but it doesn't mean $A_f^*psi$ is also in $L^2$. Though the definition of $A_f$ requires $fphi$ to be square integrable given $phiinmathrm{dom}(A_f)$
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:27










          • $begingroup$
            The adjoint is an (unbounded) operator on $L^2$. In particular, it maps into $L^2$.
            $endgroup$
            – MaoWao
            Dec 7 '18 at 12:27












          • $begingroup$
            I mean even with $A^*_fpsi=fpsi$. It is still not valid to show $A_f^*psiin L^2$
            $endgroup$
            – Apocalypse
            Dec 7 '18 at 12:28












          • $begingroup$
            Once again: If $T$ is a densely defined operator on the Hilbert space $H$, then $T^ast$ is an operator on $H$. This is just the definition of the adjoint.
            $endgroup$
            – MaoWao
            Dec 7 '18 at 12:30
















          $begingroup$
          I am not so clear about that, nothing requires $A_f^*psi$ to be in $L^2$.
          $endgroup$
          – Apocalypse
          Dec 7 '18 at 12:23




          $begingroup$
          I am not so clear about that, nothing requires $A_f^*psi$ to be in $L^2$.
          $endgroup$
          – Apocalypse
          Dec 7 '18 at 12:23












          $begingroup$
          We just know $psi$ is in the domain of $A_f^*$, but it doesn't mean $A_f^*psi$ is also in $L^2$. Though the definition of $A_f$ requires $fphi$ to be square integrable given $phiinmathrm{dom}(A_f)$
          $endgroup$
          – Apocalypse
          Dec 7 '18 at 12:27




          $begingroup$
          We just know $psi$ is in the domain of $A_f^*$, but it doesn't mean $A_f^*psi$ is also in $L^2$. Though the definition of $A_f$ requires $fphi$ to be square integrable given $phiinmathrm{dom}(A_f)$
          $endgroup$
          – Apocalypse
          Dec 7 '18 at 12:27












          $begingroup$
          The adjoint is an (unbounded) operator on $L^2$. In particular, it maps into $L^2$.
          $endgroup$
          – MaoWao
          Dec 7 '18 at 12:27






          $begingroup$
          The adjoint is an (unbounded) operator on $L^2$. In particular, it maps into $L^2$.
          $endgroup$
          – MaoWao
          Dec 7 '18 at 12:27














          $begingroup$
          I mean even with $A^*_fpsi=fpsi$. It is still not valid to show $A_f^*psiin L^2$
          $endgroup$
          – Apocalypse
          Dec 7 '18 at 12:28






          $begingroup$
          I mean even with $A^*_fpsi=fpsi$. It is still not valid to show $A_f^*psiin L^2$
          $endgroup$
          – Apocalypse
          Dec 7 '18 at 12:28














          $begingroup$
          Once again: If $T$ is a densely defined operator on the Hilbert space $H$, then $T^ast$ is an operator on $H$. This is just the definition of the adjoint.
          $endgroup$
          – MaoWao
          Dec 7 '18 at 12:30




          $begingroup$
          Once again: If $T$ is a densely defined operator on the Hilbert space $H$, then $T^ast$ is an operator on $H$. This is just the definition of the adjoint.
          $endgroup$
          – MaoWao
          Dec 7 '18 at 12:30


















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