Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $q$ ?

share|improve this question
    
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

3 Answers 3

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

share|improve this answer
    
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 en.wikipedia.org/wiki/Lacunary_function 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.

share|improve this answer

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 $L^1(U)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.