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})


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!

This entry was posted in math.CA, math.CV and tagged , . Bookmark the permalink.

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