a question about Differential quotient operator in sobolev spaces











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Define $nabla_h u=frac{tau_h u-u}{h}$, where $tau_h u=u(x_1+h,x_2,...,x_n)$.
We use $|cdot|$ to denote the norm in $L^2(R^n)$.



We have the following lemma:




Lemma 1. If $u$ belongs to $H^1(R_+^n)$, then $|nabla_hu|leq|frac{partial u}{partial x_1}|$.




Then we will have the following two statements:




  1. From $|nabla_h u|leq|frac{partial u}{partial x_1}|$, we can deduce that $tau_hu$ is a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$.


My question is:





  1. Why is $tau_hu$ a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$? I know that continuity of the operator can be deduced from the boundedness of the operator, but I don't know why the operator $tau_h u$ is bounded.



    I think that boundedness of $tau_hu$ means that there exists a constant $C$ such that
    $$supfrac{|tau_hu|_{H^1(R_+^n)}}{|u|_{L_2(R_+^n)}}leq C$$
    But I can't figure out how to prove this.








functional-analysis pde sobolev-spaces







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    Define $nabla_h u=frac{tau_h u-u}{h}$, where $tau_h u=u(x_1+h,x_2,...,x_n)$.
    We use $|cdot|$ to denote the norm in $L^2(R^n)$.



    We have the following lemma:




    Lemma 1. If $u$ belongs to $H^1(R_+^n)$, then $|nabla_hu|leq|frac{partial u}{partial x_1}|$.




    Then we will have the following two statements:




    1. From $|nabla_h u|leq|frac{partial u}{partial x_1}|$, we can deduce that $tau_hu$ is a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$.


    My question is:





    1. Why is $tau_hu$ a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$? I know that continuity of the operator can be deduced from the boundedness of the operator, but I don't know why the operator $tau_h u$ is bounded.



      I think that boundedness of $tau_hu$ means that there exists a constant $C$ such that
      $$supfrac{|tau_hu|_{H^1(R_+^n)}}{|u|_{L_2(R_+^n)}}leq C$$
      But I can't figure out how to prove this.








    functional-analysis pde sobolev-spaces







    share|cite|improve this question









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    chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      up vote
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      down vote

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      up vote
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      down vote

      favorite











      Define $nabla_h u=frac{tau_h u-u}{h}$, where $tau_h u=u(x_1+h,x_2,...,x_n)$.
      We use $|cdot|$ to denote the norm in $L^2(R^n)$.



      We have the following lemma:




      Lemma 1. If $u$ belongs to $H^1(R_+^n)$, then $|nabla_hu|leq|frac{partial u}{partial x_1}|$.




      Then we will have the following two statements:




      1. From $|nabla_h u|leq|frac{partial u}{partial x_1}|$, we can deduce that $tau_hu$ is a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$.


      My question is:





      1. Why is $tau_hu$ a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$? I know that continuity of the operator can be deduced from the boundedness of the operator, but I don't know why the operator $tau_h u$ is bounded.



        I think that boundedness of $tau_hu$ means that there exists a constant $C$ such that
        $$supfrac{|tau_hu|_{H^1(R_+^n)}}{|u|_{L_2(R_+^n)}}leq C$$
        But I can't figure out how to prove this.








      functional-analysis pde sobolev-spaces







      share|cite|improve this question









      New contributor




      chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      Define $nabla_h u=frac{tau_h u-u}{h}$, where $tau_h u=u(x_1+h,x_2,...,x_n)$.
      We use $|cdot|$ to denote the norm in $L^2(R^n)$.



      We have the following lemma:




      Lemma 1. If $u$ belongs to $H^1(R_+^n)$, then $|nabla_hu|leq|frac{partial u}{partial x_1}|$.




      Then we will have the following two statements:




      1. From $|nabla_h u|leq|frac{partial u}{partial x_1}|$, we can deduce that $tau_hu$ is a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$.


      My question is:





      1. Why is $tau_hu$ a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$? I know that continuity of the operator can be deduced from the boundedness of the operator, but I don't know why the operator $tau_h u$ is bounded.



        I think that boundedness of $tau_hu$ means that there exists a constant $C$ such that
        $$supfrac{|tau_hu|_{H^1(R_+^n)}}{|u|_{L_2(R_+^n)}}leq C$$
        But I can't figure out how to prove this.








      functional-analysis pde sobolev-spaces




      functional-analysis pde sobolev-spaces






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      edited 6 hours ago





















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      asked 7 hours ago









      chloe hj

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          Regarding the first question:



          $$sup_{uin H^1} frac{|tau_h u|_{L_2}}{|u|_{H^1}}=sup_{uin H^1} frac{|tau_h u-u+u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{|tau_h u-u|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{hcdot|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqmax(1,h)sup_{uin H^1} frac{|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leq\max(1,h)sup_{uin H^1} frac{|u|_{H^1}}{|u|_{H^1}}=max(1,h)$$



          Here I have used the triangle inequality and Lemma 1. Note that in your definition of the boundedness the norms are swapped.



          Regarding your second question, it seems to be the limit for $hto0$.






          share|cite|improve this answer








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            Regarding the first question:



            $$sup_{uin H^1} frac{|tau_h u|_{L_2}}{|u|_{H^1}}=sup_{uin H^1} frac{|tau_h u-u+u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{|tau_h u-u|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{hcdot|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqmax(1,h)sup_{uin H^1} frac{|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leq\max(1,h)sup_{uin H^1} frac{|u|_{H^1}}{|u|_{H^1}}=max(1,h)$$



            Here I have used the triangle inequality and Lemma 1. Note that in your definition of the boundedness the norms are swapped.



            Regarding your second question, it seems to be the limit for $hto0$.






            share|cite|improve this answer








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              up vote
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              Regarding the first question:



              $$sup_{uin H^1} frac{|tau_h u|_{L_2}}{|u|_{H^1}}=sup_{uin H^1} frac{|tau_h u-u+u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{|tau_h u-u|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{hcdot|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqmax(1,h)sup_{uin H^1} frac{|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leq\max(1,h)sup_{uin H^1} frac{|u|_{H^1}}{|u|_{H^1}}=max(1,h)$$



              Here I have used the triangle inequality and Lemma 1. Note that in your definition of the boundedness the norms are swapped.



              Regarding your second question, it seems to be the limit for $hto0$.






              share|cite|improve this answer








              New contributor




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                up vote
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                Regarding the first question:



                $$sup_{uin H^1} frac{|tau_h u|_{L_2}}{|u|_{H^1}}=sup_{uin H^1} frac{|tau_h u-u+u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{|tau_h u-u|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{hcdot|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqmax(1,h)sup_{uin H^1} frac{|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leq\max(1,h)sup_{uin H^1} frac{|u|_{H^1}}{|u|_{H^1}}=max(1,h)$$



                Here I have used the triangle inequality and Lemma 1. Note that in your definition of the boundedness the norms are swapped.



                Regarding your second question, it seems to be the limit for $hto0$.






                share|cite|improve this answer








                New contributor




                maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                Regarding the first question:



                $$sup_{uin H^1} frac{|tau_h u|_{L_2}}{|u|_{H^1}}=sup_{uin H^1} frac{|tau_h u-u+u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{|tau_h u-u|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqsup_{uin H^1} frac{hcdot|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leqmax(1,h)sup_{uin H^1} frac{|tfrac{partial u}{partial x_1}|_{L_2}+|u|_{L_2}}{|u|_{H^1}}leq\max(1,h)sup_{uin H^1} frac{|u|_{H^1}}{|u|_{H^1}}=max(1,h)$$



                Here I have used the triangle inequality and Lemma 1. Note that in your definition of the boundedness the norms are swapped.



                Regarding your second question, it seems to be the limit for $hto0$.







                share|cite|improve this answer








                New contributor




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                answered 6 hours ago









                maxmilgram

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