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Gamelin has the following exercise:

Let $f$ be a bounded analytic function on the upper half-plane $\{\text {Im }z>0\}$, and let $x_0 \in \Bbb R$. Assume that $f(z) \to L$ as $z \to x_0$ along a vertical line. Then $f(z) \to L$ uniformly through any cone of the form $\epsilon <\text{arg}(z-x_0)<\pi-\epsilon$.

I think this is a direct corollary from Abel's limit theorem, which reads:

Let $f$ be a bounded analytic function on the open unit disk, such that $f(x) \to L$ as $x \to 1$ for $x \in \Bbb R$. Then $f(z) \to L$ uniformly as $z \to 1$ through any cone with vertex at $1$, centered on the radius and with aperture $<\pi$.

My reasoning is that we can rotate and translate $f$ by defining $g(z)=f(-i(z+ix_0-1))$ so that $g$ is defined on the half-plane $\{\text {Re } z<1\}$ and tends to $L$ as $z \to 1$ horizontally.

Now a cone of the form $\epsilon <\text{arg}(z-x_0)<\pi-\epsilon$ in the upper half-plane corresponds to a cone with vertex at $1$ centered on the real line pointing left, so (I'm not sure) if $z \to 1$ in this cone then it must eventually be inside the open unit disk, so we can invoke Abel's limit theorem and we are done.

Is this argument sound?

Emolga
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  • Yes. The first proposition is a direct corollary of the second one. But, how do you prove the socond proposition? If you know a proof of the second proposition, please write it. – ts375_zk26 Apr 17 '17 at 22:42
  • @ts375_zk26 The second proposition is well-known, for a proof see in https://en.wikipedia.org/wiki/Abel%27s_theorem and the missing detail is in here: https://math.stackexchange.com/questions/190185/stolz-region-stuck-with-the-proof. However, is the part that I'm not sure about really valid? How would you write it in a convincing way? – Emolga Apr 17 '17 at 22:54
  • We aren't given that the sum of power series coefficients converges, so I'm not sure why Abel is being mentioned. – zhw. Apr 17 '17 at 22:56
  • The second proposition is not Abel's limit theorem. Read carefully description in en.wikipedia and the assumption of the second proposition. – ts375_zk26 Apr 17 '17 at 23:01
  • @zhw In the second highlighted claim, since the function is analytic it is also real-analytic, and since this real function has a limit at the point $1$, this limit is equal to the power series representing the function when we plug in $1$, which is the sum of the coefficients. Am I missing something? – Emolga Apr 17 '17 at 23:03
  • "the part that I'm not sure about" is really valid. But, your last comment is not correct. $$f(x)=\sum_{n=0}^\infty a_nx^n\to s (x\to 1)\implies f(1)=\sum_{n=0}^\infty a_n=s$$ doen't hold. Some additional conditions are required. If ,for instance, $na_n\to 0$ is satisfied, $\sum a_n=s$ holds. – ts375_zk26 Apr 17 '17 at 23:13
  • @Leullame It's only analytic or real analytic in the open disc. We have no information about what's happening on the boundary. – zhw. Apr 18 '17 at 00:02
  • Thank you both, I understand now what you kindly pointed. – Emolga Apr 18 '17 at 01:17

2 Answers2

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Yes, the argument is correct thanks to the geometry of the circle. Another method is to consider the family $\{f_t: t \in R^+ \}$, where $f_t(z) = f(tz)$. Since $f$ is bounded, this is a normal family. So you know that for any sequence $(t_n)$ approaching zero, $f_{t_n}$ converges to a holomorphic function $g$ locally uniformly (to see this, you can use Montel's theorem to extract sub-subsequences). Now $g = 0$ (why?).

Now let's take a compact subset $K$ as a finite-length line parallel to the real axis, say $\{x + i: -T \leq x \leq T \}$. Make $T$ large enough to intersect with the cone. Uniform convergence of $f_{t_n}$ to $g$ on $K$ is precisely a rephrasing of the problem statement.

user479334
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Yes, this argument is correct by applying a Möbius transformation that transforms the first problem to the second.

However, the second problem is not a corollary of Abel's limit theorem and is a simple consequence of Vitali's theorem on normal families.

Emolga
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