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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.

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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 $\Omega$.

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  • $\begingroup$ 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 $D_{|w_k|}(z_0) - D_{w_{k+1}}(z_0)$? $\endgroup$ – Ryan S Oct 2 '18 at 15:16

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