Let $B\subseteq \Bbb{R}^2$ be a closed ball of radius $\delta < 1$ centered at $(0,0)$. Let $f:B\to \Bbb{R}_{\geq 0}$ be real-analytic, have only one zero (at $(0,0)$) and be strictly increasing along lines that intersect the origin.
$\textbf{Claim.}$ Let $\tilde{f}(r,\theta)$ be the polar coordinate form of $f$ and let $n_{\theta}$ be the smallest non-negative integer such that $\tilde{f}^{(n_{\theta})}(0,\theta) \ne 0$. Then, $$ \inf_{\theta \in [0,2\pi)} \tilde{f}^{(n_{\theta})}(0,\theta) > 0. $$
$\textbf{Ideas.}$ As $f$ is real-analytic, it is equal to it's Taylor series $$ f(x,y) = \sum_{n,k = 0}^{\infty}c_{nk}x^ny^k, \quad c_{nk} = \frac{\partial^{n+k}}{\partial x^n \partial y^k}\frac{f(0,0)}{n!k!}. $$ Converting to polar coordinates yields $$ \tilde{f}(r,\theta) = \sum_{m=0}^{\infty}d_m(\theta) r^m, \quad d_m(\theta) = \sum_{n+k = m}c_{nk}\cos^n(\theta)\sin^k(\theta). $$ Thus, we immediately obtain that $$ \tilde{f}^{(n_{\theta})}(0,\theta) = d_{n_{\theta}}(\theta)\cdot n_{\theta}!. $$ For now we can assume that $\sup_{\theta\in [0,2\pi)}n_{\theta} < \infty$. At this point I was considering proof by contradiction. Suppose that $$ \inf_{\theta \in [0,2\pi)} \tilde{f}^{(n_{\theta})}(0,\theta) = 0. $$ Then, for all $\epsilon > 0$, there exists some $\theta_{\epsilon} \in [0,2\pi)$ such that $$ \tilde{f}^{(n_{\theta_{\epsilon}})}(0,\theta_{\epsilon}) < \epsilon \implies \frac{d_{n_{\theta_{\epsilon}}}(\theta_{\epsilon})}{n_{\theta_{\epsilon}}!}<\epsilon. $$ However, this is where I hit a wall. Any advice or counter-example would be appreciated.
