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I came across the following problem asked in a prelim exam.

Let $f$ be a function defined on the unit disk with the property that for every triplet $a, b, c$ there exists a holomorphic function $g$ such that $g$ is bounded by $1$ on the unit disk and $g(a)=f(a), g(b)=f(b)$ and $g(c)=f(c).$ Show that $f$ is holomorphic on disk and bounded by $1$.

The fact that $f$ is bounded by $1$ is trivial. Because at every point it agrees with a function which is in turn bounded by 1. But I am trying to prove the differentiablity of $f$ and I am not able to.

I was thinking to show that $f$ is differentiable at $0$ and showing that is enough to do so.

I first assume that $f(0)=0.$ To prove that $f$ is differentiable at zero, I take a sequence of points $z_n\to 0$ and obtain a sequence of holomorphic functions $g_n$ such that $f(z_n)=g_n(z_n)$. I can see that $g_n$ is normal, and hence has a convergent subsequence. I choose such a subsequence, and denote the limit by $g$. I want to show that $f’(0)=g’(0).$

I am having problem because I can get the convergence of $g_n$ only along a subsequence, and along different subsequence I may possibly have different limits.

Any suggestion is welcome.

Ravindra
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Raghav
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    I do not know the solution, but note that your $g_n$ interpolates $f$ only at two points ($0$ and $z_n$), so you are not using the full power of the assumption. – Btw, is it obvious (or easier to prove) that $f$ is continuous? – Martin R Apr 12 '20 at 06:27
  • I don't have a solution, but my first thought is to look at triangles within the unit disk. For each triangle, you have some holomorphic $g$ such that $f=g$ on the triangle's vertices. Now you can do things with the triangles, like integrate around them. One possible useful result (a version of Morera's Theorem) is that $f$ must be holomorphic if it is continuous and its integral around every triangle is $0$. Mind you, I don't know how to satisfy those hypotheses, but it's the sort of thing that I'd look at when given information about triples of points. – Toby Bartels Apr 12 '20 at 07:04
  • @Martin R, At least I do not see any easier proof which will give us the continuity. – Raghav Apr 12 '20 at 07:48
  • @Toby Bartels Just knwoing the function at 3 points doesn’t help for the integral. (At least I do not see a way to control the integral). – Raghav Apr 12 '20 at 07:49
  • Knowing that $f$ is bounded by $1$ gives you some control, although not enough to solve the problem as far as I can see. (Else I would have posted a solution.) Anyway, Martin R has found a solution that works, so never mind my idea. – Toby Bartels Apr 12 '20 at 12:22

1 Answers1

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You are on the right track. Using the full power of the assumption one can not only show that $\frac{f(z_{n_k})-f(0)}{z_{n_k} - 0}$ is convergent for some subsequence, but also that the limit is the same for all subsequences. It is a standard argument to conclude that the full sequence $\frac{f(z_{n})-f(0)}{z_{n} - 0}$ is convergent.

Fix $a \in \Bbb D$ (the unit disk) and define $$ h: \Bbb D \setminus \{ a \} \to \Bbb C, \, h(z) = \frac{f(z)-f(a)}{z-a} \, . $$ We need to show that $\lim_{z \to a} h(z)$ exists.

First consider two arbitrary sequences $(z_n)$, $(w_n)$ in $\Bbb D$ with $$ \lim_{n \to \infty} z_n = \lim_{n \to \infty} w_n = a \, . $$ For each $n$ let $g_n$ be a holomorphic function which is bounded by one and satisfies $g_n(a) = f(a)$, $g_n(z_n) = f(z_n)$, $g_n(w_n) = f(w_n$).

$(g_n)$ is a normal family and has a (locally uniformly) convergent subsequence $(g_{n_k})$. Let $g$ be the limit function. Then $g_{n_k}' \to g'$ uniformly in a neighborhood of $z=a$, and therefore $$ h(z_{n_k}) = \frac{f(z_{n_k})-f(a)}{z_{n_k} - a} = \frac{g_n(z_{n_k})-g_n(a)}{z_{n_k} - a} = \int_0^1 g_n'(a + t (z_{n_k}-a)) dt \to g'(a) $$ and similarly $$ h(w_{n_k}) \to g'(a) $$ for $k \to \infty$, i.e. $h(z_{n_k})$ and $h(w_{n_k})$ are both convergent, with the same limit.

We have therefore shown that

  • Every sequence $z_n \to a$ has a subsequence $(z_{n_k})$ such that $h(z_{n_k})$ is convergent, and
  • For any two sequences $z_n \to a$ and $w_n \to a$ such that $h(z_n)$ and $h(w_n)$ are both convergent, the limit is the same.

It follows that $\lim_{z \to a} h(z)$ exists, i.e. $f$ is differentiable at $z=a$.

Martin R
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    Thanks, Martin. I see what I was missing. I am still amazed by this problem. We do not even have continuity assumption on $f.$ It is plain magic. – Raghav Apr 12 '20 at 20:45