Inequality with analytic functions on the unit ball Let $g(z) = \sum_{n\geqslant 0} a_nz^n$ be an analytic function where $a_n$ only take values in $\{0,1\}$ (not sure if it is a necessary condition, it is just the case I'm considering).
Let $\{n\geqslant 0:a_n =1\}$ be unbounded, so $g$ is analytic only inside the unit ball $\{|z|<1\}$.
Is it true that, for all fixed $\alpha \in [0,1]$ holds
$$ |g'(\rho e(\alpha))| = O\left(\left| \frac{g(\rho e(\alpha))}{1-\rho}\right|\right),\text{ as $\rho \to 1^{-}$}$$
uniformly for $\alpha$ on the range? Here $e(\alpha) := e^{2\pi i \alpha}$.
Here's my reasoning:

Let $\alpha\in [0,1]$ fixed and take any point on the radial line with angle $2\pi i \alpha$ (i.e. choose a $0<\rho < 1$). By Cauchy's inequality, taking an open ball centered in $\rho e(\alpha)$ with radius $r$ gives
  $$ |g'(\rho e(\alpha))| \leqslant \frac{\max\{|g(z)|: z\in \mathcal{B}(\rho e(\alpha); r)\}}{r}. $$
  Given the distance from $\rho e(\alpha)$ to the rim, $r$ could be taken arbitrarily close to $1-\rho$. As the $\max$ must be near the the rim (Here's where I think is the problem!), we may take $r=r(\rho, \alpha) \approx 1-\rho$ as a smart function of $\rho$ as it tends to $1$, thus giving the inequality for some constant $M_\alpha$ on the right side.
About the uniformity, there also must be a proper way to take $r(\rho,\alpha)$ continuous on $\alpha$, so uniformity follows by compactness.

That's roughly all of my attempt in proving. To disprove I thought of taking $g(z) = \sum_{n\geqslant 0} z^{n!}$, so $g'(z) = \sum_{n\geqslant 0} n!z^{n!-1}$, and it appears to me that its poles on the boundary must be mildly wilder than those of $g$. If that is really the case, I would restrain my question to the condition $\sum_{n\leqslant x} a_n = \Omega(x^\varepsilon)$ for some $\varepsilon>0$ (the growth of the coefficients is bounded by some polynomial)
Tips and hints are totally welcome!
 A: Here are a few thoughts on the problem, too long for a comment, but which do not constitute a satisfactory answer.
First, as a consequence of the Weierstrass factorization theorem, given any countable collection $\{ z_j \} \subset \mathbb{D} = \{ \lvert z \rvert < 1 \}$ that has no accumulation point in the interior $\mathbb{D}$ and $\{ n_j \} \subset \mathbb{N}$, there is a holomorphic $f$ on $\mathbb{D}$ with zeros exactly at $z_j$ with order $n_j$.
Second, note that for any closed curve $\gamma$ on which $f$ has no zeros, the quantity
$$
\text{ind}(f \circ \gamma) = \frac{1}{2\pi i}\int_\gamma \frac{f'(z)}{f(z)} dz
$$
is the number of zeros of $f$ inside the region bounded by $\gamma$, counted with multiplicity. This is the argument principle.
Let $\gamma = C_r$ be a circle of radius $r$ centered at $0$ on which $f$ has no zeros. Then if $f$ satisfies the bound in the problem, we see that
$$
\text{ind}(f \circ C_r) \leq C(1 - \lvert r \rvert)^{-1}
$$
for all $r$ sufficiently close to $1$. We can then pick $f$ to have so many zeros that this bound does not hold. However, we cannot guarantee that $f$ has $0,1$ coefficients. The point is that any proof of the desired bound must use the coefficient properties in a fundamental way. Furthermore, the desired bound is a $\sup$ bound on $f'/f$, but from this we see that even the $L^1$ bound, which is much less strict, fails to hold in general.
Finally here is an small expansion of the comment by Andrew. If we can find a sequence of zeros $\{ z_j \}$ approaching the boundary of the disk such that each of these zeros is of order $1$ (i.e. $f'(z_j) \ne 0$), then the desired bound does not hold. My first impression is that it should be possible to find a function with the desired coefficients with this property.
