a question about Differential quotient operator in sobolev spaces












1














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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    1














    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



























      1












      1








      1







      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















      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






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 19 '18 at 7:41

























      asked Nov 19 '18 at 6:33









      chloe hj

      626




      626






















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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


























              1














              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
























                1












                1








                1






                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












                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












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 19 '18 at 7:46









                maxmilgram

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