Power series radially vanishing or exploding on their circle of convergence I'm trying to understand the behavior of two kinds of power series on the complex plane. Let's call them $f(z) = \sum_{n \in \mathbb{N}}  a_n z^n$, and suppose the radius of convergence is $1$, so that $f$ is well-defined on the open unit disk of the complex plane. 
1) In the first case, we assume that $f$ radially converges to $0$ on the circle, i.e. $$\forall \theta \in \mathbb{R}, \lim_{r \to 1^-} f(re^{i \theta}) = 0.$$ 
Does $f$ have to be zero ?
2) Now we assume that $f$ radially diverges to $+\infty$ in modulus on the circle, i.e. $$\forall \theta \in \mathbb{R}, \lim_{r \to 1^-} \lvert f(re^{i \theta})\rvert = +\infty.$$ 
Does such a function exist ?
All I found were answers involving stronger hypotheses :
1) The first statement is true if $a_n = O(\frac{1}{n})$, thanks to Littlewood's tauberian theorem, and Cantor's uniqueness theorem (as they did here).
This is also true if we actually have $\forall z_0 \in \mathbb{C}, \lvert z_0 \rvert = 1 \Rightarrow \lim_{z \to z_0} f(z) = 0$. By compactness of the closed disk, one can find a neighborhood of the circle where $f$ is arbitrary small, and apply the maximum modulus principle.
2) The function $f$ can't exist if we have : $\forall z_0 \in \mathbb{C}, \lvert z_0 \rvert = 1 \Rightarrow \lim_{z \to z_0} \lvert f(z) \rvert = + \infty$. By compactness of the closed disk, one can find a neighborhood of the circle where $\lvert f \rvert$ is greater than $1$, so that $f$ admits a finite number of zeros in the open disc (they are isolated). Thus $\frac{1}{f}$ is a meromorphic function on the open disc, and one can multiply it by a polynomial to obtain a holomorphic function $g$ satisfying : $\forall z_0 \in \mathbb{C}, \lvert z_0 \rvert = 1 \Rightarrow \lim_{z \to z_0} g(z) = 0$. Then $g$, and $\frac{1}{f}$, are zero, according to the case 1), which is impossible.
Thanks a lot.
 A: 1) The answer is yes. Let's start with a special case: Suppose $f\in H^\infty(\mathbb D)$ and $f$ has radial limit $0$ in the arc $A$ of positive length. Then $f\equiv 0.$
Proof: We can find rotations $f_1,\dots, f_n$ of $f$ such that
$$g(z) = \prod_{k=1}^n f_k(z)$$
has radial limit $0$ everywhere (we've used the boundedness of $f$ here). Now $g\in H^\infty(\mathbb D),$ hence is the Cauchy integral of its radial boundary function. Hence $g\equiv 0,$ which implies $f \equiv 0.$
Back to your problem. Let $E_n = \{t\in [0,2\pi]: |f(re^{it})| < 1 \text { for } r> 1-1/n.\}.$ Then each $E_n$ is closed and $[0,2\pi] = \cup_n E_n.$ By Baire, some $E_n$ contains an open interval, hence contains a closed interval $[a,b]$ of positive lenth. It follows that $f$ is bounded in the open sector $S=\{re^{it}: 0< r < 1, t\in (a,b)\}.$ So $f|_S \in H^\infty(S).$ Because $f$ has radial limit $0$ on an entire arc on $\partial S,$ $f$ is $0$ in $S.$ Why? By the result above. Well, we're not in the disc, we're in a sector. But some fiddling with conformal maps and a few other details will give the same result for $S.$ And of course since $f=0$ in  S  implies $f=0$ in $\mathbb D.$
2) Are there any analytic functions in $\mathbb D$ with $\lim_{r\to 1^-} |f(re^{it})| = \infty$ for all $t?$ Answer: No. You can base a proof on the ideas given for 1).
A: (Further to zhw's answer)
Improved lemma :
Suppose
\begin{align} 
&g \in H^\infty(\mathbb{D}), \\
&\Delta = \{r e^{it}, 0<r<1, t \in ]\theta_1,\theta_2[\} \text{ is an angular sector of $\mathbb{D}$}, \\
&\gamma_t : [0,1] \rightarrow \Delta \text{ is a family of continuously differentiable curves, for } t \in ]0,1[ \text{, such that :} \\  
&\qquad \lim \limits_{t \to 1}\gamma_t(0) = e^{i\theta_1}, \lim \limits_{t \to 1}\gamma_t(1) = e^{i\theta_2}, \text{ and} \lim \limits_{t \to 1} g \circ \gamma_t = 0 \text{ almost everywhere}.
\end{align}
Then $g = 0$. 
Proof :
Let's consider $\alpha = (\theta_2 - \theta_1)$, and $n = \left \lceil \frac{\alpha}{2\pi} \right \rceil $. Then we can define $\hat g\in H^\infty(\mathbb{D})$ by : 
$$ \hat g(z) = \prod_{k=0}^{n-1} g(e^{ik\alpha}z).$$
We can also define a family of continuous and piecewise-continuously differentiable closed curves $\hat \gamma_t : \mathbb{S}^1\rightarrow \mathbb{D}$ this way :
for $t \in ]0,1[$, consider the inital curve $\gamma_t$ on $]1-t, t[$, rotate its image $n$ times by the rotation of angle $\alpha$, and glue them together with straight lines. We have that $\lim \limits_{t \to 1} \hat g \circ \hat \gamma_t = 0 \text{ almost everywhere}.$
Since the winding number of $\hat \gamma_t$ around $0$ is $1$, Cauchy's integral formula and Lebegue's dominated convergence theorem implie that $\hat g = 0$, so that $g = 0$. 
In our case, we obtained $f_{|S} \in \mathbb{H}^\infty(S)$ with radial limit zero, where $S = \{r e^{it}, 0<r<1, t \in ]a, b[\}$.
Let's choose an arbitrary point $A \in S$, and denote by $\phi$ the conformal mapping from $S$ to $\mathbb{D}$ which maps $A$ to $0$. 
Now we can use the lemma with 
\begin{align} 
&g = f \circ \phi^{-1}, \\
&\Delta \text{ the angular sector of } \mathbb{D} \text{ whose extremal points are } \phi(e^{ia}) \text{ and } \phi(e^{ib}), \\
& \gamma_t = \phi \circ \eta_t : [a, b] \rightarrow \Delta,
\end{align}
where $\eta_t$ is defined by : for $t \in ]0,1[$, for $\theta \in ]a + (1-t), b - (1-t)[, \eta_t(\theta) = te^{i\theta}$, and on $[a, a + (1-t)]$ (resp. $[b - (1-t), b]$), $\eta_t$ is constant equals to $te^{a + (1-t)}$ (resp. $te^{b - (1-t)}$).
We have : $\forall \theta \in ]a, b[, \lim \limits_{t \to 1}g \circ \gamma_t(\theta) = \lim \limits_{t \to 1} f \circ \eta_t(\theta) = 0$.
Thus $g$, and $f$, are $0$.
