## Fatou’s Theorem (Complex Analysis) I

Let $F: \mathbb{D} \rightarrow \mathbb{C}$ be a holomorphic function. What conditions do we have to impose on $F$ on the boundary $\partial\mathbb{D}$ to guarantee convergence (in some sense) to boundary values? We begin by formulating a suitable form of convergence to answer this question.

We say that a function $F: \mathbb{D} \rightarrow \mathbb{C}$ has a radial limit at the point $\theta \in [-\pi,\pi]$ on the circle if the limit

$\displaystyle \lim_{{r \rightarrow 1} \atop{r < 1}} f(re^{i\theta})$

exists.

Before we give sufficient conditions, we need to review some basic results about Fourier series. For $0 \leq r < 1$ and $\theta \in [-\pi,\pi]$, we have by the geometric series formula the following identity:

$\displaystyle \sum_{n=-\infty}^{\infty}r^{\left|n\right|}e^{in\theta} = \frac{1-r^{2}}{1 - 2r\cos(\theta) + r^{2}} =: P_{r}(\theta)$

We call $P_{r}$ the Poisson kernel on the unit disk $\mathbb{D}$.

Proof. Noting that, for $n \in \mathbb{Z}^{\geq 0}$, $(re^{\pm i\theta})^{n} = r^{n}e^{\pm i n \theta}$, we obtain from the geometric series formula that

$\begin{array}{lcl} \sum_{n=-\infty}^{\infty}r^{\left|n\right|}e^{in\theta} = \sum_{n=1}^{\infty}r^{n}e^{-in\theta} + \sum_{n=0}^{\infty}r^{n}e^{in\theta} = \frac{re^{-i\theta}}{1 - re^{-i\theta}} + \frac{1}{1 - re^{i\theta}} &=& \frac{re^{-i\theta}(1 - re^{i\theta}) + (1-re^{-i\theta})}{(1-re^{-i\theta})(1-re^{i\theta})}\\ &=& \frac{re^{-i\theta} - r^{2} + 1 - re^{-i\theta}}{1 - (re^{i\theta} + re^{-i\theta}) + r^{2}}\\ &=& \frac{1-r^{2}}{1 - 2r\cos(\theta) + r^{2}} \end{array},$

since $re^{i\theta} +re^{-i\theta} = 2\text{Real}(re^{i\theta}) = 2r\cos(\theta)$. $\Box$

We use the Poisson kernel to prove the next theorem on the convergence of Fourier series. We will use (without proof of) the result that $\left\{P_{r}\right\}_{0 \leq r < 1}$ is an approximate identity.

Theorem 1. Suppose $f \in L^{1}([-\pi,\pi])$. Denote the Fourier coefficients of $f$ by $a_{n}$, $n \in \mathbb{Z}$.

1. If $a_{n} = 0$ for all $n$, then $f = 0$ a.e.
2. $\sum_{n=-\infty}^{\infty}a_{n}r^{\left|n\right|}e^{inx} \rightarrow f(x)$ for a.e. $x$, as $0 < r {\rightarrow}^{-}1$.

Proof. (1) is a consequence of (2), so we only show the latter. Changing $f(\pi)$ to $f(-\pi)$ if necessary, we may assume without loss of generality that $f(-\pi) = f(\pi)$, so that $f$ has a clear $2\pi$-periodic extension to all of $\mathbb{R}$. By the dominated convergence theorem and the above identity,

$\begin{array}{lcl}\lim_{{r \rightarrow 1} \atop {0 \leq r < 1}}\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x-y)P_{r}(y)dy &=& \lim_{{r \rightarrow 1} \atop {0 \leq r < 1}}\sum_{n=-\infty}^{\infty}\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x-y)e^{iny}dy\right)r^{\left|n\right|}\\&=& \lim_{{r \rightarrow 1} \atop {0 \leq r < 1}}\sum_{n=-\infty}^{\infty}r^{\left|n\right|}\left(\frac{1}{2\pi}\int_{x-\pi}^{x+\pi}f(y)e^{in(x-y)}dy\right)\\&=& \lim_{{r \rightarrow 1} \atop {0 \leq r < 1}}\sum_{n=-\infty}^{\infty}a_{n}r^{\left|n\right|}e^{inx} \ \text{a.e.} \end{array},$

where the penultimate equality follows from translation invariance and periodicity. Since the Poisson kernel is an approximate identity,

$\displaystyle \lim_{{r \rightarrow 1} \atop {0 \leq r}}\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x-y)P_{r}(y)dy=\lim_{{r \rightarrow 1} \atop {0 \leq r < 1}} (f \ast P_{r})(x) = f(x)$

at every Lebesgue point of $f$, hence almost everywhere. $\Box$

As above, let $a_{n}$ denote the $n^{th}$ Fourier coefficient of a function $f \in L^{1}([-\pi,\pi])$. We now prove some fundamental results for Fourier series of functions in $L^{2}([-\pi,\pi])$.

Theorem 2. Suppose $f \in L^{2}([-\pi,\pi])$.

1. (Parseval’s Identity) $\sum\left|a_{n}\right|^{2}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|f(x)\right|^{2}dx$
2. The mapping $f \mapsto (a_{n})_{n \in \mathbb{Z}}$ defines a unitary transformation from $L^{2}([-\pi,\pi])$ to $\ell^{2}(\mathbb{Z})$.
3. $\left\|f - S_{N}(f)\right\|_{L^{2}([-\pi,\pi])} \rightarrow 0$, where $S_{N}(f) := \sum_{\left|n\right| \leq N}a_{n}e^{inx}$.

Proof. Assertions (1), (2), and (3) follow from more general Hilbert space results for complete orthonormal bases. We know that $\left\{e^{inx}\right\}_{n \in \mathbb{Z}}$ is an orthonormal system, so it remains for us to completeness. Recall that a system is complete if and only if the only element orthogonal to all the elements is $0$. By assertion (1) of Theorem 1,

$\displaystyle \langle{f,e_{n}}\rangle_{L^{2}([-\pi,\pi])} := \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx = 0, \indent \forall n \in \mathbb{Z}$

implies that $f = 0$ a.e. $\Box$

Theorem 3. (Fatou) A bounded holomorphic function $F: \mathbb{D} \rightarrow \mathbb{C}$ has radial limits at almost every $\theta$.

ProofSince $F$ is holomorphic on $\mathbb{D}$, we can write $F(z) = \sum_{n=0}^{\infty}a_{n}z^{n}$, for $a_{n} \in \mathbb{C}$, where convergence holds absolutely uniformly for $z = re^{i\theta}$ with $r< 1$. For $r \in (0,1)$, define a curve $\gamma_{r} : [-\pi,\pi] \rightarrow \mathbb{C}$ by $\gamma(\theta) = re^{i\theta}$. By the uniqueness of Laurent series and the formula for Laurent coefficients,

$\begin{array}{lcl} & & a_{n} = \frac{1}{2\pi i}\int_{\gamma_{r}}\frac{F(z)}{z^{n+1}}dz = \frac{1}{2\pi i}\int_{-\pi}^{\pi}F(re^{in\theta})r^{-(n+1)}e^{-i(n+1)\theta}i(re^{i\theta})d\theta \\&\Longleftrightarrow& a_{n}r^{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}F(re^{in\theta})e^{-in\theta}d\theta\end{array}$

Note that the last integral vanishes for $n < 0$, since $F$ is holomorphic on $\mathbb{D}$. Hence, $\sum_{n=0}^{\infty}a_{n}r^{n}e^{in\theta}$ is the Fourier series of the function $\theta \mapsto F(re^{i\theta})$, for $0 \leq r < 1$ fixed.

Let $M > 0$ be such that $\left|F(z)\right| \leq M$ for all $z \in \mathbb{D}$. By Parseval’s identity,

$\displaystyle \sum_{n=0}^{\infty}\left|a_{n}\right|^{2}r^{2n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}\left|F(re^{i\theta})\right|^{2}d\theta, \indent \forall 0 \leq r < 1$

Since $F$ is dominated by $M$, hence $\sum_{n=0}^{\infty}\left|a_{n}\right|^{2}r^{2n} \leq M^{2}$, it follows from the dominated convergence theorem that $\sum_{n=0}^{\infty}\left|a_{n}\right|^{2} \leq M^{2}$. Let $F(e^{i\theta})$ denote the $L^{2}([-\pi,\pi])$ function whose Fourier coefficients $a_{n}$, $n \in \mathbb{Z}$. By assertion (2) of Theorem 1 above,

$\displaystyle \sum_{n=0}^{\infty}a_{n}r^{n}e^{in\theta} \rightarrow F(e^{i\theta}),$

for a.e. $\theta$, which completes the proof of the theorem. $\Box$

We define the Hardy space $H^{2}(\mathbb{D})$ to be the set of all holomorphic functions $F: \mathbb{D} \rightarrow \mathbb{C}$ such that

$\displaystyle \sup_{0 \leq r < 1}\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|F(re^{i\theta})\right|^{2}d\theta < \infty$

It follows from Minkowski’s inequality and basic limit properties that

$\displaystyle\left\|F\right\|_{H^{2}(\mathbb{D})} := \left(\sup_{0 \leq r < 1}\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|F(re^{i\theta})\right|^{2}d\theta\right)^{\frac{1}{2}}$

defines a norm on $H^{2}(\mathbb{D})$. We will later see that we can extend the definition of Hardy spaces on the unit disk to arbitrary $p \in (0,\infty)$.

Stay tuned for the second installment on Fatou’s theorem!