## 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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