$\textbf{Conjecture.}$ Let $B\subseteq \Bbb{R}^2$ be a closed ball centered on $(0,0)$ of radius $\delta <1$. Let $f:B\to \Bbb{R}_{\geq 0}$ be real analytic and suppose that $(0,0)$ is the only zero of $f$. Additionally, assume that $f$ is strictly increasing along lines intersecting the origin. Let $\tilde{f}(r,\theta)$ be the polar form of $f$. Then, $$ \sup_{\theta \in [0,2\pi)}\max_{r\leq \delta}\frac{\log\left(\frac{\tilde{f}(r,\theta)}{\tilde{f}(\delta,\theta)}\right)}{\log(r)}<\infty. $$ $\textbf{Ideas.}$The place where $$ \frac{\log\left(\frac{\tilde{f}(r,\theta)}{\tilde{f}(\delta,\theta)}\right)}{\log(r)} $$ can blow up is at $r = 0$. One can show that $$ \lim_{r\to 0}\frac{\log\left(\frac{\tilde{f}(r,\theta)}{\tilde{f}(\delta,\theta)}\right)}{\log(r)} = n_{\theta}, $$ where $n_{\theta}\in \Bbb{N}$ is the smallest non-negative integer such that $\tilde{f}^{(n_{\theta})}(0,\theta)\ne 0$. Moreover, one can also show that $$ \lim_{r\to \delta}\frac{\log\left(\frac{\tilde{f}(r,\theta)}{\tilde{f}(\delta,\theta)}\right)}{\log(r)} = 0. $$ For my purposes we can assume that $\sup_{\theta\in [0,2\pi)}n_{\theta} \leq N < \infty$. Now, let us assume that $$ \sup_{\theta \in [0,2\pi)}\max_{r\leq \delta}\frac{\log\left(\frac{\tilde{f}(r,\theta)}{\tilde{f}(\delta,\theta)}\right)}{\log(r)} = \infty.$$ In other words, for all $M\in \Bbb{N}$, there exists some $\theta_{M}\in [0,2\pi)$ and $r_M\leq \delta$ such that $$ \frac{\log\left(\frac{\tilde{f}(r_M,\theta_M)}{\tilde{f}(\delta,\theta_M)}\right)}{\log(r_M)}\geq M \implies \log\left(\frac{\tilde{f}(r_M,\theta_M)}{\tilde{f}(\delta,\theta_M)}\right) \leq M \log(r_M) \implies \frac{\tilde{f}(r_M,\theta_M)}{\tilde{f}(\delta,\theta_M)} \leq r_M^M. $$ Hence, if we choose $M\gg N$, then that would imply that $\tilde{f}(r,\theta_M)\sim r^M$, which would be a contradiction. Any advice would be appreciate.
