Rotation of the coordinate system for multi-index differentiations

Question

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.

Answer 1

I suspect that there is probably something way easier I just miss, that the authors alluded to, but below there's one proof of that fact.

I also believe that there must be a phrasing issue there. I will prove that given that vector $c$ and the appropriate multindex $\beta$, then by rotation one can deduce their conditions. I do believe that it is enough for the rest of their arguments later in the paper.

Fix such $c$, and $f=c_0+\sum_{i=1}^{n}c_{i}f_{i}$. Assume of-course that $\beta$ is not just the zeroth derivative. We may assume $x_0=0$.

Recall the formula $$ D^{m}_{\hat{v}}f(0) = \sum_{\lvert \alpha\rvert = m}\frac{\lvert \alpha\rvert !}{\alpha!}D_{\alpha}f(0)(\hat{v})^{\alpha}, $$ in the usual multi-index notations. Moreover, $\hat{v}$ is a unit vector.

By their assumption, there exists some multiindex $\beta$ with the derivative being non-zero. We may assume $\lvert \beta\rvert$ is minimal with this respect. We may assume $\lvert \beta\rvert=m$. Parameterizing $S^{n-1}$ in regular polar coordinates, the above formula yields a trigonometrical polynomial.

Obviously by integrating (maybe taking a double cover in various axis to make it easier), the $L^{2}$ norm squared of the polynomial is $$ \left\lVert D^{m}_{\hat{v}}f(0)\right\rVert_{L^{2}(\hat{v})}^2 = \sum_{\lvert \alpha\rvert=m}\left(\frac{\lvert \alpha\rvert!}{\alpha!}D_{\alpha}f(0)\right)^2\geq C'_{1}>0,$$ because of our assumptions and their result.

Notice we can replace $\hat{v}$ by $P\hat{v}$ for any orthogonal matrix $P$ (and essentially acheive the same $L^{2}$ bound).

For a given orthogonal matrix $P$, define $M_{f}^{P}(\epsilon) = \left\{\hat{v} \mid \left\lvert D^{m}_{P.\hat{v}}f(0)\right\rvert >\epsilon \right\}$. By our $L^{2}$ bound, $meas(M_{f}^{P}(\kappa))>0$ for small enough $\kappa>0$, where the measure is taking with respect to the (normalized) Haar measure over $S^{n-1}$. Moreover, $meas(\cup_{\kappa>0}M_{f}^P(\kappa))=1$. Hence for every $P$, there exists $\epsilon_{0}(P)>0$ (which in principle depend on $C'_{1}$) such that $meas(M_{f}^{P}(\epsilon_{0}(P))>1-1/n$, where $n$ is the dimension.

Taking $P_{1},\ldots, P_{n}$ to be the related rotation matrices, and using a union bound, one can see that $meas(M_{f}^{P_{1}}(\epsilon_{1}) \cap \cdots \cap M_{f}^{P_{n}}(\epsilon_1))>0$, for some $\epsilon_{1}=\epsilon_{1}(\epsilon_{0}(P_{1}),\ldots ,\epsilon_{0}(P_{n}))>0$.

I guess what they meant now is to continue for all the possible $\beta$'s that will appear as the appropriate multi-indices, then using a proper union bound (changing the requirement of the measure to be larger than $1-1/(n\cdot l^n)$ or so...), one can satisfy all the possible choices of multi-indices, but notice that the actual $\beta$ is dependent of $c$.