"Banach Spaces of Analytic Functions" by Kenneth Hoffman is an excellent introduction to $H^{p}$ spaces and is considered a classic mathematical analysis textbook. I was therefore surprised to find major mistakes in the proof of a central theorem ("Fatou's Theorem"). Below you will find the proofs in question (pages 79-81) followed by short explanations of their shortcomings. I will give an overview of the proofs so this can be skipped on first reading. Here $A$ is the space of continuous functions on the closed unit disk that are analytic in the interior and $Q_{r}(t)=\text{Im}(\frac{1+re^{i\theta}}{1-re^{i\theta}})$.
$\textbf{Proofs:}$
Theorem. Let $f$ be an integrable function on the circle and $$v(r,\theta)=\frac{1}{2\pi}\int_{-\pi}^\pi f(\theta-t)Q_r(t)~\mathrm{d}t.$$ If $\theta$ is any number such that the integral $$v(\theta)=-\frac{1}{2\pi}\int_{-\pi}^\pi\frac{f(\theta+t)-f(\theta-t)}{2\tan\frac{1}{2}t}~\mathrm{d}t$$ exists, then $\lim_{r\to1}v(r,\theta)=v(\theta)$.
Proof. Let $$\phi_\theta(t)=\frac{f(\theta+t)-f(\theta-t)}{2\tan\frac{1}{2}t}$$ so that we are assuming $\phi_\theta$ is integrable. Now \begin{align*} v(r,\theta)-v(\theta) &=\frac{1}{2\pi}\int_{-\pi}^\pi\phi_\theta(t)\left[1-\frac{2r\sin t\tan\frac{1}{2}t}{1-2r\cos t+r^2}\right]~\mathrm{d}t \\ &=\frac{1}{2\pi}\int_{-\pi}^\pi\phi_\theta(t)\frac{(1-r)^2}{1-2r\cos t+r^2}~\mathrm{d}t. \end{align*} Now if $$g_r(t)=\frac{(1-r)^2}{1-2r\cos t+r^2}$$ then $0<g_r(t)<1$ and $\lim_{r\to1}g_r(t)=0$, except at $t=0$. Since $\phi_\theta$ is integrable, we must have $\int\phi_\theta g_r\to0$, i.e., $$\lim_{r\to1}v(r,\theta)=v(\theta).$$
Corollary. If $f$ is differentiable at $\theta$ then $$\lim_{r\to1}v(r,\theta)=v(\theta)$$ exists. If, say, $f$ is continuously differentiable on a closed interval $\lvert\theta-\theta_0\rvert\leq\delta$, then on that interval the functions $v_r$ converge uniformly as $r\to1$.
Proof. The function $\phi_\theta$ is clearly integrable on any interval $\lvert t\rvert\geq\varepsilon>0$. If $f$ is differentiable at $\theta$, then $\phi_\theta$ is bounded as $t\to0$, so $\phi_\theta$ is integrable. If $f$ is continuously differentiable on $\lvert\theta-\theta_0\rvert\leq\delta$, then we obtain a uniform bound on $\phi_\theta$ for $\theta$ in the interval and $t$ small; it is easy to see that $v(r,\theta)$ is uniformly close to $v(\theta)$.
Theorem (Fatou). Let $K$ be a closed set of Lebesgue measure zero on the unit circle. Then there exists a function in $A$ which vanishes precisely on $K$.
Proof. Let $w$ be an extended real-valued function on the circle such that
- $w=-\infty$ on $K$, and tends continuously to $-\infty$ as $e^{i\theta}$ approaches $K$;
- $w\leq-1$ on the circle;
- $w$ is finite-valued and continuously differentiable on $C-K$;
- $w$ is integrable.
Such a $w$ can be found since $K$ has measure zero. One naive way to construct such a function is the following. Since $K$ is closed, the complement $C-K$ is the union of a countable number of disjoint open intervals (arcs) $I_n$. Let $\varepsilon_n$ be the length of $I_n$. Choose a strictly positive and continuously differentiable function $y_n$ on $I_n$ such that $y_n\leq e^{-1}$, $y_n$ tends to zero at the endpoints of $I_n$, and $$\int_{I_n}\log y_n\geq-2\varepsilon_n.$$ If we define $y$ to be zero on $K$ and $y=y_n$ on $I_n$, then $0\leq y\leq e^{-1}$; the zeros of $y$ are precisely the points of $K$; $y$ is continuous on $C$ and continuously differentiable on $C-K$; and $\log y$ is integrable. Let $w=\log y$.
Now define $$h(z)=\frac{1}{2\pi}\int_{-\pi}^\pi\frac{e^{i\theta}+z}{e^{i\theta}-z}w(\theta)~\mathrm{d}\theta.$$ Then $h$ is analytic in the open disc and $\operatorname{Re}h\leq-1$. By property 3 of $w$ and the Corollary above, $h$ is actually continuous on the complement of $K$ in the closed disc. Since $w$ tends continuously to $-\infty$ at each point of $K$, the function $$\operatorname{Re}h(r,\theta)=\frac{1}{2\pi}\int_{-\pi}^\pi w(t)P_r(\theta-t)~\mathrm{d}t$$ tends radially to $-\infty$ for each $\theta$ in $K$.
Now let $$g=\frac{1}{h}.$$ It is apparent that $g$ is in $A$, $\operatorname{Re}g\leq0$, and the zeros of $g$ on the closed disc are exactly the points of $K$. We remark that $\operatorname{Re}g=0$ exactly on $K$.
$\textbf{Errors:}$
The goal of Fatou's theorem is to construct a function $g$ in $A$ who's restriction to the boundary vanishes exactly on a given subset $K$ of the circle with measure zero. In the proof of Fatou's theorem this is done by first constructing an analytic function in the open disk, $h$, via a continuous function on the boundary, $w$. Since the complement of $K$ on the circle consists of open arcs, to show that the imaginary part, $v(r,\theta)$, of $h$ is continuous on the complement of $K$ in the closed disk, it suffices to show that $v_{r}(\theta)=v(r,\theta)$ converges uniformly as $r\rightarrow 1$ on the corresponding sector. This is the content of the corollary where $w$ will be playing the role of $f$.
$\textbf{1.}$ The biggest problem with the proof of Fatou's theorem is its reliance on the previous Corollary. Hoffman claims that continuous differentiability of $f$ on an interval $[\theta_{0}-\delta, \theta_{0}+\delta]$ implies that $v(r,\theta)$ converges uniformly to $v$ on this interval as $r\rightarrow 1$, but the proof he gives falls short. He claims in the last sentence of the proof of the corollary that we can find $\epsilon>0$ such that $\phi_{\theta}(t)$ is uniformly bounded on $[\theta_{0}-\delta, \theta_{0}+\delta]\times [-\epsilon,\epsilon]$. The idea seems clear: since $f'$ is bounded on this interval we can appeal to the mean value theorem to show that $\phi_{\theta}(t)$ is bounded. But a glance at the definition of $\phi_{\theta}(t)$ shows that this argument wont work since when $\theta=\theta_{0}\pm\delta$, $f(\theta\pm t)$ takes values outside of $[\theta_{0}-\delta, \theta_{0}+\delta]$ whenever $t\neq 0$. Now, the mean value theorem does imply that $\phi_{\theta}(t)$ is uniformly bounded for $\theta$ in a slightly smaller interval, but then we don't get the uniform convergence of $v_{r}$ on the entire interval $[\theta_{0}-\delta, \theta_{0}+\delta]$ which is needed for the proof of Fatou's theorem (where $[\theta_{0}-\delta, \theta_{0}+\delta]$ plays the role of the arcs $I_{n}$).
$\textbf{2.}$ The second issue that I see with the proof of Fatou's theorem is with the construction of $w$, which relies in turn on the construction of a continuous function $y$ on the circle. However the function $y$ he constructs need not be continuous at points of $K$ as he claims, which is necessary for $w$ to converge to $-\infty$ at points of $K$. Indeed, since the average of $\text{log}(y_{n})$ is greater than $-2$ on $I_{n}$, $y_{n}$ must take value greater than $e^{-2}$ on each $I_{n}$. But if $x\in K$, then it is possible that every neighborhood of $x$ contains an arc $I_{n}$, and hence $y$ cannot be continuous at $x$.
$\textbf{3.}$ Finally, even if a function $w$ can be constructed with the four desired properties, I do not see why radial convergence of $\text{Re}(h)$ to $-\infty$ at points $\theta\in K$ is sufficient to imply continuity of $1/h$ on the closed disk since we don't know that this convergence happens at the same rate for each point in $K$.
$\textbf{Question.}$ Considering how popular the book was (Walter Rudin calls it a "classic" in his Real and Complex Analyis) and how large the errors in this proof are, it is surpising that they were never corrected in a later edition. I would really like to know if there is any way to save these arguments to provide a correct proof of Fatou's theorem along these same lines. This would require a new argument for showing that $|v(r,\theta)-v(\theta)|$ converges uniformly on $[\theta_{0}-\delta, \theta_{0}+\delta]$ and a new way to construct a function $w$ with the desired properties.
