# Proving a holomorphic function is identically zero when its zeros form a sequence that converge in some region

Before I jump in, just want to quickly note that this post exists already, I am using the same book, and I had a very similar confusion which led me to trying to prove the theorem in a slightly different way. My question is whether or not the following proof is sufficient to establish the theorem -- the proof itself differs from the one in both Stein & Shakarchi and Rudin (see the linked post for the two book titles if it's unclear).

For ease, I'll restate the theorem:

$$\textbf{Thm:}$$ Let $$f(z)$$ be a holomorphic function defined in a region $$\Omega$$. Let $$w_k$$ denote a zero of $$f$$ in $$\Omega$$ and let the sequence $$W=\{w_k\}$$ be a sequence of zeros which converge to a limit point in $$\Omega$$. Then $$f$$ is identically zero.

$$\textbf{Pf:}$$ Let $$z_0$$ be the limit point of the sequence $$W$$. An existing theorem proves that since $$f$$ is holomorphic, it is also analytic with some positive radius of convergence when expanded at $$z_0$$. If $$f$$ is not identically zero, then there is at least one term in its expansion about $$z_0$$ which is not zero. By a separate previous theorem, we also know that the terms for the expansion at the derivatives of $$f$$ at $$z_0$$. Now, by the definition of convergence:

$$\lim_{k\to\infty}|f(z_0)-f(w_k)|=0$$

(This is the step in logic that I am unsure of): Since every of the $$f(w_k)$$ are zero, it follows that $$f(z_0)$$ is likewise zero. Hence the leading term in the expansion at $$z_0$$ must be proportional to the first derivative of $$f$$ or a higher derivative.

Next, consider the function $$f: \mathbb{C} \to \mathbb{R}$$ defined by $$z \to |f(z)|$$. Since this function has zeros at the same points in the region as $$f$$, we know that $$|f(z)|$$ has a maximum value on the interval $$(|w_k|,|w_{k+1}|)$$. Write this value as $$w'_k$$. Since the sequence $$\{w_k\}$$ converges to $$z_0$$, we have $$|w_k - w_{k+1}| \to 0$$ as $$k \to \infty$$. It follows then that $$|w'_k - w'_{k+1}|$$ also tends to $$0$$ as $$k$$ tends to infinity, where $$w'_{k+1}$$ is the value at which $$|f|$$ is max on $$(|w_{k+1}|,|w_{k+2}|)$$. But then this set also forms a sequence which converges to $$z_0$$ in the limit. But then by the same argument, $$f'(z_0)$$ must also be zero. By induction, every derivative of $$f$$ at $$z_0$$ must be zero and hence the function is identically zero in the neighborhood of $$z_0$$.

Then you can use the connectedness of the region to prove it is identically zero everywhere in $$\Omega$$.

$$\textbf{Comments:}$$ Now, I just want to mention the few spots at which I'm not 100% certain this follows:

1. Does $$f(z_0)=0$$ per the argument I made?
2. Does $$\{w'_k\}$$ also converge to $$z_0$$?
3. Does it make sense to talk about $$|f(z)|$$ being defined on $$\Omega$$ as well?

Thanks in advance for any insight offered.

## 1 Answer

1) follows from the fact that any holomorphic function is continuous. 2) is false. Your $$w_k'$$ is a positive number and it cannot converge to $$z_0$$ unless $$z_0$$ is a non-negative real number. If $$z_0=i$$, for example, then 2) fails. 3) is true. Surely $$|f(z)|$$ is well defined continuous function on $$\Omega$$.

• Could I say that the argument implies there is a complex number whose absolute value is between $|w_k|$ and $|w_{k+1}|$ which is a zero of $f'$? I.e. there is certainly a zero of $f'$ in the set $D_{|w_k|}(z_0) - D_{w_{k+1}}(z_0)$? – Ryan S Oct 2 '18 at 15:16