Identical Functions in a Domain

Question

Problem statement: If f₁(z) and f₂(z) are analytic in a domain D and equal at a sequence of points z_n in D that converges in D, show that f₁(z)=f₂(z).

I'm not clear on this - I understand that since both are analytic in D, so they both have a Taylor series in D, but the convergent set of points doesn't have to match up. I came up with what I think is a counterexample:

f₁(z)= sin(1/z), f₂(z)= -sin(1/z)

D: z = (0, 1)

z_n = 1/(n*pi), n positive integers

In this domain,

f₁(z)=f₂(z) for all z_n

In the example, I only use real numbers - not sure if that's an issue. The sum of the sequence doesn't actually converge but that's not part of the problem statement - in any case, it could be substituted with 1/(2ⁿ*pi).

Any assistance (including pointing out whatever flaws in the counterexample) would be welcome.

Answer 1

There are two issues with your counterexample: first, domains must be open, which your choice of the real interval $(0, 1)$ is not. This isn't an issue since in fact your functions are holomorphic on $\mathbb{C} \setminus \{ 0 \}$.

The bigger and much more significant issue is that the sequence $z_n$ must have an accumulation point in $D$; your proposed sequence $z_n$ has accumulation point $0$ which is not and cannot be in $D$. The proof involves the Taylor expansion around the accumulation point so this hypothesis is crucial.

Answer 2

I suppose that, in your example, $D=\Bbb C\setminus\{0\}$. But then $\lim_{n\to\infty}\frac1{n\pi}=0\notin D$. So, you did not create a counterexample; remember that the statement has the hypothesis that $\lim_{n\to\infty}z_n\in D$.