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Let $\mathbf{f} = (f_1,\dotsc, f_n)$ be a $C^l$-map from an open subset $U$ of $\mathbb R^d$ to $\mathbb R^n$, and let $x_0 \in U$ be such that $\mathbb R^n$ is spanned by partial derivatives of $f$ at $x_0$ of order $l$.

Let $f=c_0+ \sum_{i=1}^{n} c_i f_i$ with $\mathbf{c}=(c_0,\dotsc, c_n)$ being of Euclidean norm $1$. From the nondegeneracy assumption it follows that there exists a constant $C_1 > 0$ such that for any $\mathbf{c}$ with $\lVert\mathbf{c}\rVert = 1$ one can find a multi-index $\beta$ with the sum of components $\lvert\beta\rvert = k < l$ and

$$\left\lvert\sum_{i=1}^n c_i \partial_{\beta} f_i(x_0)\right\rvert = \lvert\partial_{\beta} f(x_0)\rvert \ge C_1.\tag{$*$}\label{star}$$

(The assertion above can be proved by contradiction. Suppose not then for any $C_1>0$, there exists $\mathbf{c}$ and $\lvert\beta\rvert\le l$, and $\lvert\partial_{\beta} f(x_0)\rvert<C_1$ which means $\partial_\beta f(x_0)=0$ but this contradicts to the non-degeneracy of $f$. $\mathbf{c}$ gives a linear dependency relation.)

Now here is my question:

I wonder how to show that assuming \eqref{star}, with an appropriate rotation of the coordinate system around $x_0$, one can guarantee that $\lvert\partial_i^k f(x_0)\rvert \ge C_2$ for all $i=1,\dotsc, d$ and some positive $C_2$ independent of $\mathbf c$.

Here notation $\partial_i$ means $\frac{\partial}{\partial x_i}$ and if $\beta= (k_1,\dotsc k_n) \in \mathbb N^n$, then $\partial_{\beta}f= \partial_1^{k_1}\dotsm \partial_n^{k_n} f$.

This is hard to prove for me even for the dimension two case.

Source of the question: Kleinbock and Margulis, "Flows on homogeneous spaces and Diophantine approximation on manifolds", 1998 Annals of Math at the top of page 10, the proof of the proposition 3.4.

Let $\mathbf{f} = (f_1,\dotsc, f_n)$ be a $C^l$-map from an open subset $U$ of $\mathbb R^d$ to $\mathbb R^n$, and let $x_0 \in U$ be such that $\mathbb R^n$ is spanned by partial derivatives of $f$ at $x_0$ of order $l$.

Let $f=c_0+ \sum_{i=1}^{n} c_i f_i$ with $\mathbf{c}=(c_0,\dotsc, c_n)$ being of Euclidean norm $1$. From the nondegeneracy assumption it follows that there exists a constant $C_1 > 0$ such that for any $\mathbf{c}$ with $\lVert\mathbf{c}\rVert = 1$ one can find a multi-index $\beta$ with the sum of components $\lvert\beta\rvert = k < l$ and

$$\left\lvert\sum_{i=1}^n c_i \partial_{\beta} f_i(x_0)\right\rvert = \lvert\partial_{\beta} f(x_0)\rvert \ge C_1.\tag{$*$}\label{star}$$

(The assertion above can be proved by contradiction. Suppose not then for any $C_1>0$, there exists $\mathbf{c}$ and $\lvert\beta\rvert\le l$, and $\lvert\partial_{\beta} f(x_0)\rvert<C_1$ which means $\partial_\beta f(x_0)=0$ but this contradicts to the non-degeneracy of $f$. $\mathbf{c}$ gives a linear dependency relation.)

Now here is my question:

I wonder how to show that assuming \eqref{star}, with an appropriate rotation of the coordinate system around $x_0$, one can guarantee that $\lvert\partial_i^k f(x_0)\rvert \ge C_2$ for all $i=1,\dotsc, d$ and some positive $C_2$ independent of $\mathbf c$.

Here notation $\partial_i$ means $\frac{\partial}{\partial x_i}$ and if $\beta= (k_1,\dotsc k_n) \in \mathbb N^n$, then $\partial_{\beta}f= \partial_1^{k_1}\dotsm \partial_n^{k_n} f$.

This is hard to prove for me even for the dimension two case.

Source of the question: Kleinbock and Margulis, "Flows on homogeneous spaces and Diophantine approximation on manifolds", 1998 Annals of Math at the top of page 10, the proof of the proposition 3.4.

Let $\textbf{f} = (f_1,\dots, f_n)$ be a $C^l$-map from an open subset $U$ of $\mathbb R^d$ to $\mathbb R^n$, and let $x_0 \in U$ be such that $\mathbb R^n$ is spanned by partial derivatives of $f$ at $x_0$ of order $l$.

Let $f=c_0+ \sum_{i=1}^{n} c_i f_i$ with $\textbf{c}=(c_0,\dots, c_n)$ being of Euclidean norm $1$. From the nondegeneracy assumption it follows that there exists a constant $C_1 > 0$ such that for any $\textbf{c}$ with $\|\textbf{c}\| = 1$ one can find a multi-index $\beta$ with the sum of components $|\beta| = k < l$ and

$$|\sum_{i=1}^n c_i \partial_{\beta} f_i(x_0)| = |\partial_{\beta} f(x_0)| \ge C_1. (*)$$

(The assertion above can be proved by contradiction. Suppose not then for any $C_1>0$, there exists $\textbf{c}$ and $|\beta|\le l$, and $|\partial_{\beta} f(x_0)|<C_1$ which means $\partial_\beta f(x_0)=0$ but this contradicts to the non-degeneracy of $f$. $\textbf{c}$ gives a linear dependency relation)

Now here is my question:

I wonder how to show that assuming (*), with an appropriate rotation of the coordinate system around $x_0$, one can guarantee that $|\partial_i^k f(x_0)| \ge C_2$ for all $i=1,\dots, d$ and some positive $C_2$ independent of $c$.

Here notation $\partial_i$ means $\frac{\partial}{\partial x_i}$ and if $\beta= (k_1,\dots k_n) \in \mathbb N^n$, then $\partial_{\beta}f= \partial_1^{k_1}\cdots \partial_n^{k_n} f$.

This is hard to prove for me even for the dimension two case.

Source of the question: (Kleinbock and Margulis 1998 Annals of Math) https://arxiv.org/pdf/math/9810036.pdf#page=10# at the top, the proof of the proposition 3.4

Let $\mathbf{f} = (f_1,\dotsc, f_n)$ be a $C^l$-map from an open subset $U$ of $\mathbb R^d$ to $\mathbb R^n$, and let $x_0 \in U$ be such that $\mathbb R^n$ is spanned by partial derivatives of $f$ at $x_0$ of order $l$.

Let $f=c_0+ \sum_{i=1}^{n} c_i f_i$ with $\mathbf{c}=(c_0,\dotsc, c_n)$ being of Euclidean norm $1$. From the nondegeneracy assumption it follows that there exists a constant $C_1 > 0$ such that for any $\mathbf{c}$ with $\lVert\mathbf{c}\rVert = 1$ one can find a multi-index $\beta$ with the sum of components $\lvert\beta\rvert = k < l$ and

$$\left\lvert\sum_{i=1}^n c_i \partial_{\beta} f_i(x_0)\right\rvert = \lvert\partial_{\beta} f(x_0)\rvert \ge C_1.\tag{$*$}\label{star}$$

(The assertion above can be proved by contradiction. Suppose not then for any $C_1>0$, there exists $\mathbf{c}$ and $\lvert\beta\rvert\le l$, and $\lvert\partial_{\beta} f(x_0)\rvert<C_1$ which means $\partial_\beta f(x_0)=0$ but this contradicts to the non-degeneracy of $f$. $\mathbf{c}$ gives a linear dependency relation.)

Now here is my question:

I wonder how to show that assuming \eqref{star}, with an appropriate rotation of the coordinate system around $x_0$, one can guarantee that $\lvert\partial_i^k f(x_0)\rvert \ge C_2$ for all $i=1,\dotsc, d$ and some positive $C_2$ independent of $\mathbf c$.

Here notation $\partial_i$ means $\frac{\partial}{\partial x_i}$ and if $\beta= (k_1,\dotsc k_n) \in \mathbb N^n$, then $\partial_{\beta}f= \partial_1^{k_1}\dotsm \partial_n^{k_n} f$.

This is hard to prove for me even for the dimension two case.

Source of the question: Kleinbock and Margulis, "Flows on homogeneous spaces and Diophantine approximation on manifolds", 1998 Annals of Math at the top of page 10, the proof of the proposition 3.4.

Let $\textbf{f} = (f_1,\dots, f_n)$ be a $C^l$-map from an open subset $U$ of $\mathbb R^d$ to $\mathbb R^n$, and let $x_0 \in U$ be such that $\mathbb R^n$ is spanned by partial derivatives of $f$ at $x_0$ of order $l$.

Let $f=c_0+ \sum_{i=1}^{n} c_i f_i$ with $\textbf{c}=(c_0,\dots, c_n)$ being of Euclidean norm $1$. From the nondegeneracy assumption it follows that there exists a constant $C_1 > 0$ such that for any $\textbf{c}$ with $\|\textbf{c}\| = 1$ one can find a multi-index $\beta$ with the sum of components $|\beta| = k < l$ and

$$|\sum_{i=1}^n c_i \partial_{\beta} f_i(x_0)| = |\partial_{\beta} f(x_0)| \ge C_1. (*)$$

(The assertion above can be proved by contradiction. Suppose not then for any $C_1>0$, there exists $\textbf{c}$ and $|\beta|\le l$, and $|\partial_{\beta} f(x_0)|<C_1$ which means $\partial_\beta f(x_0)=0$ but this contradicts to the non-degeneracy of $f$. $\textbf{c}$ gives a linear dependency relation)

Now here is my question:

I wonder how to show that assuming (*), with an appropriate rotation of the coordinate system around $x_0$, one can guarantee that $|\partial_i^k f(x_0)| \ge C_2$ for all $i=1,\dots, d$ and some positive $C_2$ independent of $c$.

Here notation $\partial_i$ means $\frac{\partial}{\partial x_i}$ and if $\beta= (k_1,\dots k_n) \in \mathbb N^n$, then $\partial_{\beta}f= \partial_1^{k_1}\cdots \partial_n^{k_n} f$.

This is hard to prove for me even for the dimension two case.

Source of the question: https://arxiv.org/pdf/math/9810036.pdf#page=10# at the top, the proof of the proposition 3.4

Let $\textbf{f} = (f_1,\dots, f_n)$ be a $C^l$-map from an open subset $U$ of $\mathbb R^d$ to $\mathbb R^n$, and let $x_0 \in U$ be such that $\mathbb R^n$ is spanned by partial derivatives of $f$ at $x_0$ of order $l$.

Let $f=c_0+ \sum_{i=1}^{n} c_i f_i$ with $\textbf{c}=(c_0,\dots, c_n)$ being of Euclidean norm $1$. From the nondegeneracy assumption it follows that there exists a constant $C_1 > 0$ such that for any $\textbf{c}$ with $\|\textbf{c}\| = 1$ one can find a multi-index $\beta$ with the sum of components $|\beta| = k < l$ and

$$|\sum_{i=1}^n c_i \partial_{\beta} f_i(x_0)| = |\partial_{\beta} f(x_0)| \ge C_1. (*)$$

(The assertion above can be proved by contradiction. Suppose not then for any $C_1>0$, there exists $\textbf{c}$ and $|\beta|\le l$, and $|\partial_{\beta} f(x_0)|<C_1$ which means $\partial_\beta f(x_0)=0$ but this contradicts to the non-degeneracy of $f$. $\textbf{c}$ gives a linear dependency relation)

Now here is my question:

I wonder how to show that assuming (*), with an appropriate rotation of the coordinate system around $x_0$, one can guarantee that $|\partial_i^k f(x_0)| \ge C_2$ for all $i=1,\dots, d$ and some positive $C_2$ independent of $c$.

Here notation $\partial_i$ means $\frac{\partial}{\partial x_i}$ and if $\beta= (k_1,\dots k_n) \in \mathbb N^n$, then $\partial_{\beta}f= \partial_1^{k_1}\cdots \partial_n^{k_n} f$.

This is hard to prove for me even for the dimension two case.

Source of the question: (Kleinbock and Margulis 1998 Annals of Math) https://arxiv.org/pdf/math/9810036.pdf#page=10# at the top, the proof of the proposition 3.4