Estimate for non-differentiable implicit function theorem












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$begingroup$


Let $F(x,s)$ be a continuous function $F:mathbb R^mtimesmathbb R^ntomathbb R^m$ such that $nabla_xF$ is a Holder $C^alpha$-function, say $frac12<alpha<1$.



Suppose $F(0,0)=0$ and suppose $nabla_xF(x,s)$ is a invertible matrix for every $(x,s)$.



By non-differentiable implicit function theorem mention in Wikipedia, locally there is a unique function $phi:Usubsetmathbb R^mtomathbb R^n$ near $0$, such that $F(phi(t),t)=0$.



My question is, does $phiin C^alpha$? The non-differentiable IFT only guarantee $phi$ to be continuous, but not something further.



And for the distributional derivative $nablaphi(t)=-(F_x^{-1}cdot F_s)|_{(phi(t),t)}$, the product make sense as a $C^{alpha-1}$ distribution when $alpha>frac12$. But the composition of $C^{alpha-1}circ C^alpha$ is not sure to be well-defined






real-analysis implicit-function-theorem holder-spaces







share|cite|improve this question









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    1












    $begingroup$


    Let $F(x,s)$ be a continuous function $F:mathbb R^mtimesmathbb R^ntomathbb R^m$ such that $nabla_xF$ is a Holder $C^alpha$-function, say $frac12<alpha<1$.



    Suppose $F(0,0)=0$ and suppose $nabla_xF(x,s)$ is a invertible matrix for every $(x,s)$.



    By non-differentiable implicit function theorem mention in Wikipedia, locally there is a unique function $phi:Usubsetmathbb R^mtomathbb R^n$ near $0$, such that $F(phi(t),t)=0$.



    My question is, does $phiin C^alpha$? The non-differentiable IFT only guarantee $phi$ to be continuous, but not something further.



    And for the distributional derivative $nablaphi(t)=-(F_x^{-1}cdot F_s)|_{(phi(t),t)}$, the product make sense as a $C^{alpha-1}$ distribution when $alpha>frac12$. But the composition of $C^{alpha-1}circ C^alpha$ is not sure to be well-defined






    real-analysis implicit-function-theorem holder-spaces







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $F(x,s)$ be a continuous function $F:mathbb R^mtimesmathbb R^ntomathbb R^m$ such that $nabla_xF$ is a Holder $C^alpha$-function, say $frac12<alpha<1$.



      Suppose $F(0,0)=0$ and suppose $nabla_xF(x,s)$ is a invertible matrix for every $(x,s)$.



      By non-differentiable implicit function theorem mention in Wikipedia, locally there is a unique function $phi:Usubsetmathbb R^mtomathbb R^n$ near $0$, such that $F(phi(t),t)=0$.



      My question is, does $phiin C^alpha$? The non-differentiable IFT only guarantee $phi$ to be continuous, but not something further.



      And for the distributional derivative $nablaphi(t)=-(F_x^{-1}cdot F_s)|_{(phi(t),t)}$, the product make sense as a $C^{alpha-1}$ distribution when $alpha>frac12$. But the composition of $C^{alpha-1}circ C^alpha$ is not sure to be well-defined






      real-analysis implicit-function-theorem holder-spaces







      share|cite|improve this question









      $endgroup$




      Let $F(x,s)$ be a continuous function $F:mathbb R^mtimesmathbb R^ntomathbb R^m$ such that $nabla_xF$ is a Holder $C^alpha$-function, say $frac12<alpha<1$.



      Suppose $F(0,0)=0$ and suppose $nabla_xF(x,s)$ is a invertible matrix for every $(x,s)$.



      By non-differentiable implicit function theorem mention in Wikipedia, locally there is a unique function $phi:Usubsetmathbb R^mtomathbb R^n$ near $0$, such that $F(phi(t),t)=0$.



      My question is, does $phiin C^alpha$? The non-differentiable IFT only guarantee $phi$ to be continuous, but not something further.



      And for the distributional derivative $nablaphi(t)=-(F_x^{-1}cdot F_s)|_{(phi(t),t)}$, the product make sense as a $C^{alpha-1}$ distribution when $alpha>frac12$. But the composition of $C^{alpha-1}circ C^alpha$ is not sure to be well-defined






      real-analysis implicit-function-theorem holder-spaces




      real-analysis implicit-function-theorem holder-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Feb 2 at 22:31









      yaolidingyaoliding

      599312




      599312






















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          I just assume $phi$ is already continuous due to the non-differentiable IFT. Remain to check the Holder condition.



          Use $F(phi(t),t)=F(phi(t'),t')=0$, we have
          $$|phi(t)-phi(t')|lefrac1{inf|(nabla_x F)^{-1}|}|F(phi(t),t)-F(phi(t'),t)|=frac1{inf|(nabla_x F)^{-1}|}|F(phi(t'),t')-F(phi(t'),t)|lefrac{[F(phi(t'),cdot)]_alpha}{inf|(nabla_x F)^{-1}|}|t-t'|^alphalefrac{[F]_alpha}{inf|(nabla_x F)^{-1}|}|t-t'|^alpha$$



          The problem is still annoying if we consider the Zygmund-1 function class, namely those functions $f$ satisfies $|f(x)+f(y)-2f(frac{x+y}2)|le C|x-y|$. I believe they are still true. But the estimate may require using mean value theorem, where we should replace the first inequality into equality by choosing suitable coefficients.






          share|cite|improve this answer









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            $begingroup$

            I just assume $phi$ is already continuous due to the non-differentiable IFT. Remain to check the Holder condition.



            Use $F(phi(t),t)=F(phi(t'),t')=0$, we have
            $$|phi(t)-phi(t')|lefrac1{inf|(nabla_x F)^{-1}|}|F(phi(t),t)-F(phi(t'),t)|=frac1{inf|(nabla_x F)^{-1}|}|F(phi(t'),t')-F(phi(t'),t)|lefrac{[F(phi(t'),cdot)]_alpha}{inf|(nabla_x F)^{-1}|}|t-t'|^alphalefrac{[F]_alpha}{inf|(nabla_x F)^{-1}|}|t-t'|^alpha$$



            The problem is still annoying if we consider the Zygmund-1 function class, namely those functions $f$ satisfies $|f(x)+f(y)-2f(frac{x+y}2)|le C|x-y|$. I believe they are still true. But the estimate may require using mean value theorem, where we should replace the first inequality into equality by choosing suitable coefficients.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              I just assume $phi$ is already continuous due to the non-differentiable IFT. Remain to check the Holder condition.



              Use $F(phi(t),t)=F(phi(t'),t')=0$, we have
              $$|phi(t)-phi(t')|lefrac1{inf|(nabla_x F)^{-1}|}|F(phi(t),t)-F(phi(t'),t)|=frac1{inf|(nabla_x F)^{-1}|}|F(phi(t'),t')-F(phi(t'),t)|lefrac{[F(phi(t'),cdot)]_alpha}{inf|(nabla_x F)^{-1}|}|t-t'|^alphalefrac{[F]_alpha}{inf|(nabla_x F)^{-1}|}|t-t'|^alpha$$



              The problem is still annoying if we consider the Zygmund-1 function class, namely those functions $f$ satisfies $|f(x)+f(y)-2f(frac{x+y}2)|le C|x-y|$. I believe they are still true. But the estimate may require using mean value theorem, where we should replace the first inequality into equality by choosing suitable coefficients.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                I just assume $phi$ is already continuous due to the non-differentiable IFT. Remain to check the Holder condition.



                Use $F(phi(t),t)=F(phi(t'),t')=0$, we have
                $$|phi(t)-phi(t')|lefrac1{inf|(nabla_x F)^{-1}|}|F(phi(t),t)-F(phi(t'),t)|=frac1{inf|(nabla_x F)^{-1}|}|F(phi(t'),t')-F(phi(t'),t)|lefrac{[F(phi(t'),cdot)]_alpha}{inf|(nabla_x F)^{-1}|}|t-t'|^alphalefrac{[F]_alpha}{inf|(nabla_x F)^{-1}|}|t-t'|^alpha$$



                The problem is still annoying if we consider the Zygmund-1 function class, namely those functions $f$ satisfies $|f(x)+f(y)-2f(frac{x+y}2)|le C|x-y|$. I believe they are still true. But the estimate may require using mean value theorem, where we should replace the first inequality into equality by choosing suitable coefficients.






                share|cite|improve this answer









                $endgroup$



                I just assume $phi$ is already continuous due to the non-differentiable IFT. Remain to check the Holder condition.



                Use $F(phi(t),t)=F(phi(t'),t')=0$, we have
                $$|phi(t)-phi(t')|lefrac1{inf|(nabla_x F)^{-1}|}|F(phi(t),t)-F(phi(t'),t)|=frac1{inf|(nabla_x F)^{-1}|}|F(phi(t'),t')-F(phi(t'),t)|lefrac{[F(phi(t'),cdot)]_alpha}{inf|(nabla_x F)^{-1}|}|t-t'|^alphalefrac{[F]_alpha}{inf|(nabla_x F)^{-1}|}|t-t'|^alpha$$



                The problem is still annoying if we consider the Zygmund-1 function class, namely those functions $f$ satisfies $|f(x)+f(y)-2f(frac{x+y}2)|le C|x-y|$. I believe they are still true. But the estimate may require using mean value theorem, where we should replace the first inequality into equality by choosing suitable coefficients.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 3 at 6:38









                yaolidingyaoliding

                599312




                599312






























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