Fatou’s Theorem (Complex Analysis) II

We now give some examples of holomorphic functions F: \mathbb{D} \rightarrow \mathbb{C} which fail to have radial limits. First, consider the function holomorphic F_{0}: \mathbb{D} \rightarrow \mathbb{C} defined by

\displaystyle F_{0}(z) := \frac{1}{(1-z)^{i}}

where we take the principal branch with \arg(z) \in [-\pi,\pi). I claim that F_{0} is bounded on \mathbb{D}, but \lim_{r \rightarrow 1} F_{0}(r) does not exist. Observe that, for z \in \mathbb{D}, -\frac{\pi}{2} < \arg(1-z) < \frac{\pi}{2}, so that

\displaystyle \left|F_{0}(re^{i\theta})\right| = \left|\frac{1}{(1 -z)^{i}}\right| = \left|\frac{1}{e^{i[\log\left|r\right| + i\arg(1-z)]}}\right| = \left|\frac{1}{e^{-\arg(1-z)}e^{i\log\left|r\right|}}\right| < \frac{1}{e^{-\frac{\pi}{2}}} = e^{\frac{\pi}{2}}

For the second assertion, observe that

\displaystyle \left|F_{0}(r)\right| = \left|\frac{1}{e^{i\log\left|r\right|}}\right| = 1,

but since

\displaystyle F_{0}(r) = \begin{cases} e^{(2n+1)\pi} = -1 & {r = e^{-(2n+1)\pi}, n \in \mathbb{Z}} \\ e^{2n\pi} = 1 & {r = e^{-2n\pi}, n \in \mathbb{Z}}\end{cases}

we see that F_{0}(r) oscillates between 1 and -1 infinitely often as r \rightarrow 1.

We now give an example of a holomorphic function F: \mathbb{D} \rightarrow \mathbb{C} which fails to have radial limits on a dense subset of \mathbb{R}. Let \left\{\alpha_{j}\right\}_{j=1}^{\infty} be an enumeration of the rationals. By Weierstrass’ theorem, choosing \left|\delta\right| < 1, we can define a holomorphic function F: \mathbb{D} \rightarrow \mathbb{C} by

\displaystyle F(z) := \sum_{j=1}^{\infty}\delta^{j}F_{0}(ze^{-i\alpha_{j}}), \indent z \in \mathbb{D}

I claim that \lim_{r \rightarrow 1} F(re^{i\theta}) does not exist for \theta = \alpha_{j}, for all j. Indeed, we showed above that F_{0}(r) = F(re^{i\alpha_{j}}e^{-i\alpha_{j}}) oscillates between 1 and -1, hence \delta^{j}F_{0}(r) oscillates between -\delta^{j} and \delta^{j} infinitely often. So

\begin{array}{lcl}\liminf_{r \rightarrow 1} \text{Re} F(z) = -\delta^{j} + \liminf_{r \rightarrow 1} \text{Re} \sum_{n \neq j}\delta^{n}F_{0}(ze^{-i\alpha_{n}}) &<& \delta^{j} + \liminf_{r \rightarrow 1} \text{Re} \sum_{n \neq j}\delta^{n}F_{0}(ze^{-i\alpha_{n}})\\ &\leq& \limsup_{r \rightarrow 1} \text{Re} F(z)\end{array}

We now give examples of holomorphic functions F: \mathbb{D} \rightarrow \mathbb{C} which have radial limits nowhere. The following results can be found in David G. Cantor’s note A Simple Construction of Analytic Functions without Radial Limits.

We first prove a Tauberian theorem, which we will use to give a wide class of holomorphic functions on \mathbb{D} without radial limits.

Let n_{1} < n_{2} <\cdots be an increasing sequence of positive integers such that

\displaystyle \frac{r_{k}^{n_{k+1}}}{1-r_{k}} \leq \frac{1}{2^{k}},    \forall k \in \mathbb{Z}^{\geq 1}

where r_{k} is the positive n_{k}^{th} root of 1-2^{-k}. Clearly, r_{k} < 1 for all k \in \mathbb{Z}^{\geq 1}. Observe that, for k \leq j,

\displaystyle 1 - r_{j}^{n_{k}} \leq 1 - r_{j}^{n_{j}} = 2^{-j}

We can construct such a sequence by induction. Set n_{1} := 1 and r_{1} := 1 - 2^{-1}. If we choose n_{2} := 2, then

\displaystyle \frac{r_{1}^{n_{2}}}{1 - r_{1}} = \frac{2^{-2}}{1 - 2^{-1}} = \frac{1}{2}

Suppose we have chosen positive integers n_{1} < \cdots < n_{k} and r_{1}, \cdots, r_{k-1} such that

\displaystyle \frac{r_{j}^{n_{j+1}}}{1-r_{j}} \leq \frac{1}{2^{j}},     \displaystyle \forall 1 \leq j \leq k-1

where r_{j}^{n_{j}} = 1-2^{-j}. Let r_{k} be the positive solution of r_{k}^{n_{k}} = 1 - 2^{-k}. Since 0 < r_{k} < 1, by the archimedean property of the reals, we can choose a sufficiently large positive integer n_{k+1} > n_{k} so that

\displaystyle r_{k}^{n_{k+1}} \leq \frac{1}{2^{k}}\left(1 - r_{k}\right)

Lemma 1. Suppose (a_{j})_{j=1}^{\infty} is a bounded sequence in \mathbb{C} satisfying

\displaystyle \lim_{{r \rightarrow 1} \atop {0 < r < 1}}\sum_{k=1}^{\infty}a_{k}r^{n_{k}} = s \in \mathbb{C},

where (n_{k})_{k=1}^{\infty} is as above. Then \sum_{k=1}^{\infty}a_{k} = s.

Proof. Choose A > 0 so that \left|a_{k}\right| \leq A for all k \in \mathbb{Z}^{\geq 1}. Then

\displaystyle \left|\sum_{k=1}^{j}a_{k} - \sum_{k=1}^{\infty}a_{k}r_{j}^{n_{k}}\right| \leq \left|\sum_{k=1}^{j}a_{k}(1-r_{j}^{n_{k}})\right| + \sum_{k=j+1}^{\infty}\left|a_{k}\right|r_{j}^{n_{k}} \leq A\frac{j}{2^{j}} + A\sum_{k=j+1}^{\infty}r_{j}^{n_{k}}

We need an estimate for the second term. By construction of the sequences (n_{k}) and (r_{k}), we have that

\displaystyle \frac{1}{2^{j}} \geq \frac{r_{j}^{n_{j+1}}}{1-r_{j}} = r_{j}^{n_{j}}\sum_{k=0}^{\infty}r_{j}^{k} \geq r_{j}^{n_{j}}\sum_{k=j+1}^{\infty}r_{j}^{n_{k}}

Hence,

\displaystyle \left|\sum_{k=1}^{j}a_{k} - \sum_{k=1}^{\infty}a_{k}r_{j}^{n_{k}}\right| \leq A\frac{j}{2^{j}} + A\frac{1}{2^{j}} \rightarrow 0, \ j \rightarrow \infty

\Box

Theorem 2. Let (b_{k})_{k=1}^{\infty} be a bounded sequence in \mathbb{C} not satisfying \lim_{k \rightarrow \infty} b_{k} = 0. If (n_{k})_{k=1}^{\infty} is a sequence of positive integers as above, then f(z) := \sum_{k=1}^{\infty}b_{k}z^{n_{k}} defines a holomorphic function on \mathbb{D} and f(z) has radial limits nowhere.

Proof. That f is holomorphic in \mathbb{D} is immediate from the boundedness of (b_{k}) and Weierstrass’ theorem. Suppose f has a radial limit at z = e^{i\theta}. a_{k} := e^{ik\theta}b_{k} defines a bounded sequence, so by our Tauberian theorem,

\displaystyle \lim_{{r \rightarrow 1} \atop { 0 \leq r < 1}}f(re^{i\theta}) = \lim_{{r\rightarrow 1} \atop {0 \leq r < 1}} \sum_{k=1}^{\infty}b_{k}e^{ik\theta}r^{n_{k}} = s \Longrightarrow \sum_{k=1}^{\infty}b_{k}e^{ik\theta} = s

But then \lim_{k \rightarrow \infty} \left|b_{k}e^{ik\theta}\right| = \lim_{k \rightarrow \infty} \left|b_{k}\right| = 0, which is a contradiction. \Box

Theorem 3. Let \phi: [0,r) \rightarrow \mathbb{R} be a continuous, strictly positive function, satisfying \lim_{r \rightarrow 1^{-}}\phi(r) = \infty. Let (b_{k})_{k=1}^{\infty} be a bounded sequence in \mathbb{C} not satisfying \lim_{k \rightarrow \infty}b_{k} = 0. Then there exists an increasing sequence of integers (n_{k})_{k=1}^{\infty}, such that f(z) = \sum_{k=1}^{\infty}b_{k}z^{n_{k}} is holomorphic in \mathbb{D}, has radial limits nowhere, and \sup_{\left|z\right| =r } \left|f(z)\right| \leq \phi(r).

ProofWe construct the sequence (n_{k})_{k=1}^{\infty} inductively. I claim that we can choose n_{1} \in \mathbb{Z}^{\geq 1} sufficiently large so that

\displaystyle \left|b_{1}\right|r^{n_{1}} \leq 2^{-1}\phi(r), \indent \forall 0 \leq r < 1

Indeed, for any n \in \mathbb{Z}^{\geq 1}, the function f_{n}(r) := 2\left|b_{1}\right|\frac{r^{n}}{\phi(r)} is continuous and bounded on [0,1) and \lim_{r \rightarrow 1^{-}}f(r) = 0. By Weierstrass’ extreme value theorem, f_{n} attains a maximum at some r_{n} \in (0,1). By the compactness of [0,1], (r_{n})_{n=1}^{\infty} has a convergent subsequence r_{n_{k}} \rightarrow r_{0} \in [0,1]. For \epsilon > 0 very small, we have for all n_{k} sufficiently large that \left|r_{0}-r_{n_{k}}\right| < \epsilon and \phi(r) \geq 2\left|b_{1}\right| for all r \geq r_{0} - \epsilon. Hence,

\displaystyle 2\left|b_{1}\right|\frac{r^{n_{k}}}{\phi(r)} \leq 2\left|b_{1}\right|\frac{r_{n_{k}}^{n_{k}}}{\phi(r_{n_{k}})} \leq r_{n_{k}}^{n_{k}} \leq 1, \indent \forall 0 \leq r < 1

For the inductive step, suppose we are given n_{k}. By a completely analogous argument, we can choose n_{k+1} > n_{k} sufficiently large so that

\displaystyle \left|b_{1}\right|r^{n_{k+1}} \leq 2^{-k-1}\phi(r), \indent \forall 0 \leq r < 1

and so that n_{k+1} satisfy the conditions of the Tauberian theorem above. Hence,

\displaystyle f(z) := \sum_{k=1}^{\infty}b_{k}z^{n_{k}}, \indent z \in \mathbb{D}

is holomorphic in \mathbb{D} and has radial limits nowhere. Furthermore,

\displaystyle \sup_{\left|z\right| = r} \left|f(z)\right| \leq \sum_{k=1}^{\infty}\left|b_{k}\right|r^{n_{k}} \leq \sum_{k=1}^{\infty}2^{-k}\phi(r) = \phi(r), \indent \forall 0 \leq r < 1

\Box

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