Answer: The Zeros That Refuse to Accumulate

Key Idea / Intuition

The Identity Theorem says: if the zeros of a non-zero analytic function accumulate at an interior point of the domain, then the function is identically zero. The sequence $z_n = 1 - 1/n$ accumulates at $z = 1$, which is on the boundary of $\mathbb{D}$, not inside it — so the Identity Theorem doesn't apply, and in fact a non-trivial function can vanish on this sequence. For the second case, zeros can accumulate anywhere on the boundary, as long as they don't cluster inside, so a non-zero analytic function can vanish on an arbitrary boundary-accumulating sequence — but subject to the Blaschke condition.

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Formal Proof / Solution

Case 1: $z_n = 1 - 1/n$, can $f(z_n) = 0$ for all $n$?

Answer: Yes, this is possible.

The accumulation point of $\{z_n\}$ is $z = 1$, which lies on the boundary $\partial \mathbb{D}$, not inside $\mathbb{D}$. The Identity Theorem requires an accumulation point in the domain of analyticity. Since $z=1 \notin \mathbb{D}$, the theorem gives no contradiction.

Explicit example: The Blaschke product. A sequence $\{a_n\} \subset \mathbb{D}$ is the zero set of a bounded analytic function on $\mathbb{D}$ if and only if the Blaschke condition holds: $\sum_{n=1}^\infty (1 - |a_n|) < \infty.$

For $a_n = 1 - 1/n$, we compute: $\sum_{n=1}^\infty (1 - |a_n|) = \sum_{n=1}^\infty \frac{1}{n} = \infty.$

So the Blaschke condition fails for this sequence! This means no bounded analytic function has exactly these zeros. However, an unbounded analytic function can still vanish on this sequence. For instance, consider:

$f(z) = \exp\!\left(-\frac{1}{1-z}\right).$

This is analytic on $\mathbb{D}$ (the argument $-1/(1-z)$ has real part $\to -\infty$ as $z \to 1$ along the real axis), not identically zero, and in fact $f(z_n) = e^{-n} \neq 0$. So this particular $f$ doesn't work directly, but the key point remains:

A cleaner explicit example with zeros on the sequence $\{1 - 1/n\}$: consider the function

$f(z) = \sin\!\left(\frac{\pi}{1-z}\right).$

This is analytic on $\mathbb{D}$, not identically zero, and $f(z_n) = \sin(n\pi) = 0$ for all $n \geq 1$. ✓

The zeros $z_n = 1 - 1/n$ accumulate at the boundary point $1$, not at any interior point, so the Identity Theorem is not violated.

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Why the Identity Theorem is the right tool

Identity Theorem: If $f$ is analytic on a connected open set $U$ and the zero set $\{f = 0\}$ has an accumulation point inside $U$, then $f \equiv 0$ on $U$.

The key word is inside. Boundary accumulation is not controlled by analyticity of $f$ at that point.

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Case 2: General sequence $w_n$ with $|w_n| \to 1$

Answer: Also possible, subject to the Blaschke condition.

If the sequence satisfies the Blaschke condition $\sum(1 - |w_n|) < \infty$, then the Blaschke product

$B(z) = \prod_{n=1}^\infty \frac{|w_n|}{w_n} \cdot \frac{w_n - z}{1 - \overline{w_n} z}$

converges uniformly on compact subsets of $\mathbb{D}$, is analytic and bounded ($|B| \leq 1$), and vanishes exactly on $\{w_n\}$.

If the Blaschke condition fails, then no bounded analytic function can have this zero set. However, unbounded analytic functions may still exist with such zeros (as in Case 1 above).

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Summary Table

| Situation | Accumulation point | Possible? | Reason | |---|---|---|---| | $z_n = 1 - 1/n$ | $z=1 \in \partial\mathbb{D}$ | Yes | Identity Thm doesn't apply | | $w_n \to \partial\mathbb{D}$, Blaschke holds | boundary | Yes | Blaschke product works | | Zeros accumulate at interior point | $z_0 \in \mathbb{D}$ | No | Identity Theorem |

The elegant takeaway: analyticity has perfect memory inside the domain, but no control on the boundary.