I am reading the paper Nodal sets of eigenfunctions of the Laplacian on Surfaces by Donnelly and Fefferman available here. I have a problem understanding Lemma 5.10. To my understanding, what follows is the content of the Lemma 5.10

Let $\mathbb{D}$ be the unit disc in $\mathbb{R}^2$ centered at the origin. Fix $m \in \mathbb{N}$. Then there is a constant $C_m >0$ such that for any smooth $g :\mathbb{D} \rightarrow \mathbb{R}$ satisfying the following condition \begin{align*} \max_{|\alpha| \leq m} \left| \frac{\partial^{\alpha}g}{\partial z^{\alpha}}(0) \right| \geq 1 \end{align*} there are a pair of perpendicular directions $v_1$ and $v_2$ and some $k \leq m$ (here $v_1, v_2$ and $k$ can possibly depend on the function $g$) such that the following holds \begin{align*} |\partial_{v_1}^{k}g(0)| \geq C_m~\mbox{and}~|\partial_{v_2}^{k}g(0)| \geq C_m \end{align*}

Here are my questions regarding this Lemma.

  1. Firstly, is my understanding of the statement of Lemma 5.10 correct? More specifically, is it claimed that there is a universal constant $C_m$ (independent of the function $g$)?

  2. If the answer to my first question is Yes, I cannot really see how the proof provided in the paper establishes existence of the universal constant $C_m$.

Thanks in advance!


1 Answer 1


$\newcommand{\al}{\alpha} \newcommand{\G}{\mathcal{G}} \newcommand{\PP}{\mathcal{P}}$ You are missing condition (ii) in formula (5.9) in that paper. That formula is \begin{equation} \begin{aligned} &\text{(i)}\quad \sup_{|\al|\le c_4} \Big| \frac{\partial^{\al}g}{\partial z^{\al}}(0) \Big|\ge b_6, \\ &\text{(ii)}\quad \sup_{|z|<1/2}\sup_{|\al|\le c_4+1} \Big| \frac{\partial^{\al}g}{\partial z^{\al}}(z)\Big|\le b_7. \end{aligned} \tag{5.9} \end{equation} It appears that, in the paper, the $a_i$'s, $b_i$'s, and $c_i$'s denote positive real constants, which may depend only on such constants previously mentioned in the paper.

Then Lemma 5.10 there seems to say that for some nonnegative integer $k\le c_4$ and some perpendicular unit vectors $v_1$ and $v_2$ we have \begin{align*} |\partial_{v_1}^{k}g(0)|\ge b_{11}~\mbox{and}~|\partial_{v_2}^{k}g(0)|\ge b_{11}. \end{align*}

This is indeed true, with $b_{11}>0$ depending only on $c_4,b_6,b_7$. Indeed, let $\G$ be the set of all smooth functions $g$ satisfying (5.9), and let $\PP$ be the set of the Taylor polynomials (at $0$) of order $\lfloor c_4\rfloor$ of the functions $g\in\G$. The set $\PP$ is obviously compact in the natural topology on $\PP$. So, there exists \begin{align} \mu&:=\min_{g\in\G}\max_{k,v_1,v_2}(|\partial_{v_1}^{k}g(0)|\wedge|\partial_{v_2}^{k}g(0)|) \\ & =\min_{P\in\PP}\max_{k,v_1,v_2}(|\partial_{v_1}^{k}P(0)|\wedge|\partial_{v_2}^{k}P(0)|)\in[0,\infty), \end{align} where $\max_{k,v_1,v_2}$ is over all nonnegative integers $k\le c_4$, and all perpendicular unit vectors $v_1$ and $v_2$.

Let $P$ be a minimizer for $\min_{P\in\PP}$. By condition (5.9)(i), $P$ is a nonzero polynomial of degree $\le c_4$ and hence has a a nonvanishing term of order $k\le c_4$. So, $\partial_v^{k}P(0)=0$ only for finitely many unit vectors $v$. So, we can find perpendicular unit vectors $v_1$ and $v_2$ such that $|\partial_{v_1}^{k}P(0)|\wedge|\partial_{v_2}^{k}P(0)|>0$. Thus, $\mu>0$. It remains to let $b_{11}:=\mu>0$, which depends only on $c_4,b_6,b_7$.

  • $\begingroup$ Thanks for the clarification. Makes sense now. I misunderstood the statement before Lemma 5.10 which said "We easily see that the following reformulation of (5.9)(i) holds". But do you think we can do away with Condition 5.9(ii)? Reason for thinking this might be possible is that it is true when $c_4 = 1$: If $\max\{|\partial_1 g(0)|, |\partial_2 g(0)|\} > b_6$, then taking the two directions to be the ones which are 45 degress apart from the $\nabla g(0)$ works. $\endgroup$
    – April
    Oct 21, 2019 at 17:23
  • $\begingroup$ Continuation of my previous comment: If $\max\{|\partial_1 g(0)|, |\partial_2 g(0)|\} > b_6$, then if $\theta$ is such that $\nabla g(0) = |\nabla g(0)| e^{i \theta}$, then taking $v_1 = e^{i(\theta + \pi/4)}$ and $v_2 = e^{i(\theta - \pi/4)} $ gives both $|\partial_{v_1} g(0)|$ and $|\partial_{v_2} g(0)| \geq b_6 / \sqrt{2}$ $\endgroup$
    – April
    Oct 21, 2019 at 17:31
  • 1
    $\begingroup$ @April : Condition (5.9)(i) is on the maximum of the partial derivatives, not just of the highest order $\le c_4$, but of all orders up through the highest order. $\endgroup$ Oct 21, 2019 at 17:34
  • $\begingroup$ Yes, I understand that. My point is when $c_4 = 1$, it looks like (the argument I provided in the comments) we do not need condition 5.9(ii) which is an upper bound on the the $L^{\infty}$ norm of the function and appropriate derivatives. This made me think if condition 5.9(ii) is really essential. But I fully understand and agree with the proof you provided. $\endgroup$
    – April
    Oct 21, 2019 at 17:38

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