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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 $q$ ?

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If I were you, I would study the equivalent (locally) problem of boundary behavior of harmonic functions on the upper half plane. Then you have a very convenient representation of the functions via the Poisson kernel; you can prescribe an arbitrary trace (e.g. a positive function belonging only to some Lp spaces but not others) and you get a positive harmonic function which converges to the trace monotonically – Piero D'Ancona Aug 14 '11 at 9:11

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 $q=e^{i\pi\varphi}$.

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I don't get that $+\infty$ ... what is the meaning of a sequence of complex numbers having $+\infty$ as a limit? – Samuele Aug 13 '11 at 22:36
@Samuele Yes, I made a mistake here, thanks. I fixed it. – Andrew Aug 14 '11 at 6:09
See for more background. – Emil Jeřábek Aug 16 '11 at 10:52

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.

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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 $L^1(U)$.

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