Proof of the Lindelöf theorem related to the radial limit of an analytic function in the unit disc Hi I am looking for the proof of this theorem here by Lindelöf:
"Suppose $\Gamma$ is a curve with parameter interval $[0,1]$, such that $|\Gamma(t)| < 1$ if $t < 1$ and $\Gamma(1)=1$. If $g \in H^{\infty}$ and $\lim_{t \to 1} g(\Gamma(t))=L$ then g has radial limit $L$ at $1$."
I tried Google and every result I've found was either about the Phragmén-Lindelöf theorem about analytic functions on a strip, the Lindelöf theorem about covering or the Picard-Lindelöf theorem. Typing Lindelöf theorem radial limit didn't help either because it sent me to research articles about extensions of the theorem. Can anyone please tell me where to find the proof of this theorem that is not Walter Rudin's Real and Complex Analysis? Thank you.
 A: Since you stated in a comment that a clarification of the proof would also be acceptable, let's walk through the proof Rudin gives:

Assume $\lvert g\rvert < 1$, $L = 0$, without loss of generality. Let $\varepsilon > 0$ be given. There exists $t_0 < 1$ so that, setting $r_0 = \operatorname{Re} \Gamma(t_0)$, we have $$\lvert g(\Gamma(t))\rvert < \varepsilon \quad \text{and} \quad \operatorname{Re} \Gamma(t) > r_0 > \frac{1}{2}\tag{2}$$ as soon as $t_0 < t < 1$.

There's not much going on so far, we normalise the situation so that $\lvert g\rvert < 1$ and $L = 0$ by subtracting a constant and scaling (if necessary), so that the formulae in the following are simpler to write and read. It is clear that such modifications do not influence the limiting behaviour.
Then we restrict our attention to a tail of the curve $\Gamma$ that stays close to $1$ and on which $\lvert g\rvert$ is small, where smallness is defined by the arbitrary given $\varepsilon > 0$ (which we may assume to be $< 1$, for otherwise there's nothing left to show).
In the following, we show that $\lvert g\rvert$ is also small on the outer part of the radius ending in $1$, that $\lvert g(r)\rvert \leqslant \varepsilon^{1/4}$ for all $r_0 < r < 1$. For that, we choose an arbitrary such $r$, and define a helper region and a helper function depending on $r$:

Pick $r,\, r_0 < r < 1$. Define $h$ in $\Omega = D(0;1) \cap D(2r;1)$ by $$h(z) = g(z) \overline{g(\overline{z})} g(2r-z) \overline{g(2r-\overline{z})}.\tag{3}$$ Then $h\in H(\Omega)$ and $\lvert h\rvert < 1$. We claim that $$\lvert h(r)\rvert < \varepsilon.\tag{4}$$ Since $h(r) = \lvert g(r)\rvert^4$, the theorem follows from $(4)$.

The region $\Omega$ is symmetric with respect to the real axis and the line $x = r$, so for $z\in \Omega$ we have $\overline{z}, 2r-z, 2r-\overline{z}\in\Omega \subset D(0;1)$, so $h$ is well-defined by $(3)$, holomorphic in $\Omega$ as a product of holomorphic functions, and $h$ is real and non-negative on $\Omega\cap\mathbb{R}$. Further, $h(2r-z) = h(z)$ and $h(\overline{z}) = \overline{h(z)}$, and by that we obtain a bound for $\lvert h\rvert$ on a curve that is symmetric with respect to the real axis and the line $x = r$ from the bound for $\lvert g\rvert$ on the tail of the curve $\Gamma$.

To prove $(4)$, let $E_1 = \Gamma([t_1,1])$, where $t_1$ is the largest $t$ for which $\operatorname{Re} \Gamma(t) = r$, let $E_2$ be the reflection of $E_1$ in the real axis, and let $E$ be the union of $E_1\cup E_2$ and its reflection in the line $x = r$. Then $(2)$ and $(3)$ imply that $$\lvert h(z)\rvert < \varepsilon\quad \text{if}\quad z\in \Omega \cap E.\tag{5}$$

We take the tail of $\Gamma$ to the right of $x = r$, and construct the curve $E$ by reflecting that tail in the real axis, and the line $x = r$ as well as in the point $r$. Thus $E$ is a closed curve in $\Omega \cup \{ 1,2r-1\}$ consisting of four pieces, and on each of the pieces, except for the points $1$ and $2r-1$ outside $\Omega$ one of the factors of $h$ is $g(\Gamma(t))$ or $\overline{g(\Gamma(t))}$ for some $t\in [t_1,1)$, hence less than $\varepsilon$ in absolute value, while the other factors are less than $1$ in absolute value by the bound on $\lvert g\rvert$, hence $(5)$ holds. But $h$ is not defined at the two points of $E$ outside $\Omega$, therefore we need another helper function. Since $h$ is bounded, we obtain a continuous function on $E$ by multiplying $h$ with continuous functions on $\overline{\Omega}$ that vanish at $1$ resp. at $2r-1$ and are holomorphic in $\Omega$, and bounded in absolute value by $1$.

Pick $c > 0$, define $$h_c(z) = h(z)(1-z)^c(2r-1-z)^c\tag{6}$$ for $z\in \Omega$, and put $h_c(1) = h_c(2r-1) = 0$.

Now we have a function $h_c$ that is holomorphic in $\Omega$, continuous on $E$, vanishes at $1$ and $2r-1$ and is less than $\varepsilon$ in absolute value on $E$.
Up to here, you seem to have understood everything, based on your comment, so I apologise for the long-windedness, but it may help other readers.

If $K$ is the union of $E$ and the bounded components of the complement of $E$, then $K$ is compact, $h_c$ is cotinuous on $K$, holomorphic in the interior of $K$, and $(5)$ implies that $\lvert h_c\rvert < \varepsilon$ on the boundary of $K$. Since the construction of $E$ shows that $r\in K$, the maximum modulus theorem implies that $\lvert h_c(r)\rvert < \varepsilon$. Letting $c \to 0$, we obtain $(4)$.

So let's come to the final part, about which you had questions.

I don't understand the claims about $K$ in the proof (please refer to the text). Why is $K$ compact?

The curve $E$ is compact, hence its complement is open, and has one unbounded component - the set $U = \{ z : \lvert z\rvert > 1\}$ is a connected (open) subset of $\mathbb{C}\setminus E$, and its complement is bounded, hence every unbounded component of $\mathbb{C}\setminus E$ must intersect $U$, and by connectedness contain $U$, thus there is only one unbounded component. All bounded components of $\mathbb{C}\setminus E$ are contained in the closed unit disk, hence $K$ is bounded (it is contained in the closed unit disk). The complement of $K$ is the unbounded component of $\mathbb{C}\setminus E$, hence open, so $K$ is closed. By the Heine-Borel theorem, $K$ is compact.

How do you know that $K$ contains a connected region

We do not know that, that may not be the case. If $\Gamma(t) = t$ for example, $K$ is just a closed interval on the real line. It can also happen for other curves $\Gamma$ that $K = E$, but in that case, there is nothing to show, since the desired bound is already known on $E$. Only if $\mathbb{C}\setminus E$ has a bounded component do we need the maximum modulus principle to deduce the desired bound from the bound on $\partial K$. But that is the generic case.

and that the boundary of that region is a subset of $E$,

If $V = \overset{\Large\circ}{K} \neq \varnothing$, then $\partial V \subset \overline{V}\subset K$ since $K$ is closed. And, since $V$ is open, $\partial V \cap V = \varnothing$. To see that $\partial V \subset E$, we must therefore see that $K\setminus E$ contains no boundary point of $V$. But $K\setminus E$ is the union of components of $\mathbb{C}\setminus E$, hence an open set, so $K\setminus E \subset V$, whence $(K\setminus E)\cap \partial V = \varnothing$.

and that $r$ is inside that region?

We don't know that $r$ lies in the interior of $K$ - it need not - but we know that $r$ lies in $K$. If $\Gamma(t_1) = r$, then $r\in E$ and we're done. Otherwise, let $W$ be the component of $\mathbb{C}\setminus E$ containing $r$. We need to show that $W$ is not the unbounded component.
If you know some topology, the proof is straightforward using the topological definition of winding number. Since $A = \{\Gamma(t_1)\} \cup \{ z\in \mathbb{C} : \operatorname{Re} z > r\}$ is convex, $\Gamma\lvert_{[t_1,1]}$ is homotopic in $A$, with fixed endpoints, to the straight line segment connecting $\Gamma(t_1)$ and $1$. Reflecting that homotopy, we find that the closed curve whose trace is $E$ is homotopic (as closed curves) in $\mathbb{C}\setminus \{r\}$ to the polygonal closed curve $[\Gamma(t_1), 1, \overline{\Gamma(t_1)}, 2r-1,\Gamma(t_1)]$, whose winding number around $r$ is $1$ or $-1$, depending on the sign of $\operatorname{Im} \Gamma(t_1)$. Since homotopic closed curves have the same winding number, it follows that the winding number of $E$ around $r$ is $\pm 1$, but on the unbounded component of $\mathbb{C}\setminus E$, the winding number is $0$.
If $\Gamma$ is a well-behaved curve, we can use the same argument for the winding number as it is defined in complex analysis, but unless $\Gamma$ is rectifiable, using the integral definition of the winding number is hairy.
I could offer a complicated proof avoiding the topological machinery, but unless requested, I will abstain.
