Boundary behavior of a holomorphic function on $D$ ? Hi, I have two related questions.
$D$ = open init disk in the complex plane $C$.
A.   Let $f: D \to C $ be a holomorphic function. Then is it possible that $\forall   q \in S^1$,there exists a sequence $ (x_n) \to q$ such that $f(x_n)\to \infty $  ?
B. Let $1\le p \le \infty$.  Fix   $p$.   Let $f:D\to C$ be a holomorphic function. Then must there exists a point $q\in S^1$ such that in some neighborhood $U $ of $q$ in $D$, $f \in L^p(U)$.i.e., on the contrary, is it possible to have a holomorphic function $f$ not belonging to $ L^p(U)      \forall  q \in S^1$ and $\forall $ neighborhood $U$ of $q$ ?
 A: For the first question the answer is positive. Consider the function $f(x)=\sum_{n=1}^\infty x^{2^n}$. Then $f(x^2)=f(x)-x$. Since $\lim_{x\to1-0}f(x)=+\infty$ then the same is true as $x=re^{i \pi\psi }\to q=e^{i \pi\psi }$ where $\psi$ is any binary rational point (of the form $k/2^n$, $k\in \mathbb Z$, $n\in \mathbb Z_+$). Approximating now an arbitrary $\varphi$ by such numbers we will have the statement for any $q=e^{i\pi\varphi}$. 
A: Suppose f is in $L^p(U)$. Then $f$ has a trace on the boundary that is in $W^{-1/p,p}$.
You can conclude this, for instance, from the trace theorem for divergence free vector fields: If f=u+iv, then the vector fields (u,-v) and (v,u) are both divergence free, so their normal components have a trace on the boundary.
On the other hand, you can prescribe a distribution on the boundary that is not in $W^{-1/p,p}$ on any subinterval of the boundary, and then find a harmonic function which takes on these boundary values.
A: Consider $g(z) = 1/(1-z)^2 = \sum_{n=0}^\infty (n+1) z^n$, which satisfies $\int_D |g(z)| \ dx\ dy = \infty$.  Let $S_{N,M}(z) = \sum_{N \le n < M} (n+1) z^n$ and $B_r(q) = \{z \in D: |z - q| < r\}$. For any $r > 0$ and any $N$ and $R$, since $\int_{B_r(1)} |S_{N,\infty}(z)|\ dx\ dy = \infty$, for $\epsilon > 0$ sufficiently small we have
$\int_{B_r(1) \cap B_{1-\epsilon}(0)} ||S(N,\infty)(z)|\ dx\ dy > 2 R$, and we can take $M$ large enough that $\int_{B_r(1) \cap B_{1-\epsilon}(0)} |S(N,M)(z)|\ dx\ dy > R$.
Enumerate pairs $(r, q)$ where $1/r$ is a positive integer and $q = \exp(i t)$ where $t$ is rational as $(r_k, q_k)$.  Now define $f(z) = \sum_{k=1}^\infty S_{N_k, N_{k+1}}(z/q_k)$ where $N_k$ is an increasing sequence and $\epsilon_k>0$ a decreasing sequence, defined inductively so that 
$$\int_{B_{r_k}(1) \cap B_{1-\epsilon_k}(0)} |S_{N_k,N_{k+1}}(z)| \, dx \, dy> k + 
\sum_{j < k} \int_D |S_{N_j,N_{j+1}}(z)|\, dx \, dy$$
and $\sum_{n \ge N_{k+1}} (n+1) (1-\epsilon_k)^n < 1/{2 \pi}$. The sum converges, uniformly on compact sets, to an analytic function on $D$ with
\begin{eqnarray*}
&\int_{B_{r_k}(q_k) \cap B_{1 - \epsilon_k}(0)}  |f(z)|\, dx \, dy \cr
&\ge \int_{B_{r_k}(q_k) \cap B_{1 - \epsilon_k}(0)} (|S_{N_k,N_{k+1}}(z/q_k)| - 
\sum_{j < k} |S_{N_j,N_{j+1}}(z/q_k)| - \sum_{j \ge N_{k+1}} (n+1) (1-\epsilon_k)^j)\, dx\, dy\cr
&\ge k - 1\cr
\end{eqnarray*}
Since every neighbourhood $U$ of any $q \in S^1$ contains infinitely many $B_{r_k}(q_k)$, $f$ is not in $L^1(U)$.
