Mixing
Estimate for non-differentiable implicit function theorem
$begingroup$
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
asked Feb 2 at 22:31
yaolidingyaoliding
599312
$endgroup$
add a comment |
$begingroup$
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
asked Feb 2 at 22:31
yaolidingyaoliding
599312
$endgroup$
add a comment |
$begingroup$
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
asked Feb 2 at 22:31
yaolidingyaoliding
599312
$endgroup$
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
asked Feb 2 at 22:31
yaolidingyaoliding
599312
asked Feb 2 at 22:31
yaolidingyaoliding
599312
asked Feb 2 at 22:31
yaolidingyaoliding
599312
asked Feb 2 at 22:31
yaolidingyaoliding
599312
asked Feb 2 at 22:31
yaolidingyaoliding
599312
599312
add a comment |
add a comment |
1 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.
answered Feb 3 at 6:38
yaolidingyaoliding
599312
$endgroup$
add a comment |
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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.
answered Feb 3 at 6:38
yaolidingyaoliding
599312
$endgroup$
add a comment |
$begingroup$
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.
answered Feb 3 at 6:38
yaolidingyaoliding
599312
$endgroup$
add a comment |
$begingroup$
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.
answered Feb 3 at 6:38
yaolidingyaoliding
599312
$endgroup$
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.
answered Feb 3 at 6:38
yaolidingyaoliding
599312
answered Feb 3 at 6:38
yaolidingyaoliding
599312
answered Feb 3 at 6:38
yaolidingyaoliding
599312
answered Feb 3 at 6:38
yaolidingyaoliding
599312
599312
add a comment |
add a comment |
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