Limit point (Tag D-TFSPT)
Let \(X\) be a topological space, let \(S\subseteq X\), and let \(x\in X\). The point \(x\) is a limit point of \(S\) if every neighborhood of \(x\) contains a point of \(S\) other than \(x\).
Let \(K\) be a compact subset of a topological space \(X\). Every infinite subset \(S\subseteq K\) has a limit point in \(K\).
Suppose no point of \(K\) is a limit point of \(S\). Then each \(x\in K\) has an open neighborhood \(U_x\) with \(U_x\cap S\subseteq\{x\}\). The sets \(U_x\) cover \(K\), so finitely many \(U_{x_1},\ldots,U_{x_m}\) cover \(K\), and then \(S\subseteq\{x_1,\ldots,x_m\}\) is finite.
Equivalent conditions for a holomorphic function to vanish identically (Tag T-SVF2W)
Let \(\Omega\subseteq\CC\) be a nonempty connected open set and let \(f\) be holomorphic on \(\Omega\). The following are equivalent.
\(f(z)=0\) for all \(z\in\Omega\).
There are \(z_0\in\Omega\) and a sequence \((z_k)_{k\geq1}\) of distinct points of \(\Omega\) with \(z_k\to z_0\) and \(f(z_k)=0\) for all \(k\).
There is \(z_0\in\Omega\) with \(f^{(n)}(z_0)=0\) for all \(n\geq0\).
Two holomorphic functions on a connected open set that agree on a set with a limit point in the set are equal.
(Using power series and topology)
\(1\implies 2\): Choose \(z_0\in\Omega\) and \(r>0\) with \(D_r(z_0)\subseteq\Omega\), and take the distinct points \(z_k \coloneqq z_0 + r/(k+1)\), which are zeros of \(f\) converging to \(z_0\).
\(2\implies 3\): Expand \(f\) in a power series \(\sum_k c_k(z-z_0)^k\) about the accumulation point \(z_0\in\Omega\). Suppose some \(c_m\neq 0\), and take \(m\) minimal. Then \[ f(z) = \sum_{k\geq m}c_k (z-z_0)^k = (z-z_0)^m \sum_{k\geq m}c_k (z-z_0)^{k-m} \coloneqq (z-z_0)^m g(z) ,\] where \(g\) is holomorphic near \(z_0\) and \(g(z_0) = c_m \neq 0\). By continuity \(g\) is nonzero on a neighborhood \(U \ni z_0\), so \(f\) has no zeros in \(U\sm\ts{z_0}\). Since the \(z_k\) are distinct and converge to \(z_0\), infinitely many of them lie in \(U\sm\ts{z_0}\), a contradiction. So every \(c_k = f^{(k)}(z_0)/k!\) vanishes.
\(3\implies 1\): Let \(U \coloneqq \ts{w\in \Omega\st f^{(k)}(w) = 0 \text{ for all }k }\), which contains \(z_0\). For \(w\in U\), the power series of \(f\) about \(w\) has all coefficients \(f^{(k)}(w)/k! = 0\), so \(f\equiv 0\) on a disc about \(w\) and that disc lies in \(U\); hence \(U\) is open. Each set \(\ts{f^{(k)} = 0}\) is closed in \(\Omega\) by continuity of \(f^{(k)}\), so their intersection \(U\) is closed in \(\Omega\). Since \(\Omega\) is connected and \(U\) is nonempty, \(U=\Omega\) and \(f\equiv 0\).
(A limit point outside the domain, and a disconnected domain)
The function \(\sin(1/z)\) on \(\CC\sm\ts{0}\) vanishes at the distinct points \(1/(k\pi) \to 0\) and is not identically zero; the limit point \(0\) is not in the domain. On the disconnected open set \(\DD\cup D_1(3)\), the function equal to \(0\) on \(\DD\) and to \(1\) on \(D_1(3)\) is holomorphic, vanishes on \(\DD\), and is not identically zero.
Identity theorem, with a two-variable version (Tag T-OFMGU)
Let \(\Omega\subseteq\CC\) be a connected open set and let \(S\subseteq\Omega\) have a limit point in \(\Omega\).
If \(f\) and \(g\) are analytic on \(\Omega\) and \(f(z)=g(z)\) for all \(z\in S\), then \(f(z)=g(z)\) for all \(z\in\Omega\).
If \(F\colon\Omega\times\Omega\to\CC\) is holomorphic in each variable separately and \(F(z,w)=0\) for all \(z,w\in S\), then \(F(z,w)=0\) for all \(z,w\in\Omega\).
For (i), the zeros of the analytic function \(f-g\) have a limit point in the connected open set \(\Omega\), so \(f-g\equiv0\). For (ii), fix \(w\in S\); the function \(z\mapsto F(z,w)\) is holomorphic on \(\Omega\) and vanishes on \(S\), so by (i) it vanishes on \(\Omega\). Now fix \(z\in\Omega\); the function \(w\mapsto F(z,w)\) is holomorphic on \(\Omega\) and vanishes on \(S\), so by (i) it vanishes on \(\Omega\).
(Transferring a real identity to the plane)
Since \(\sin^2(z)+\cos^2(z) = 1\) on \(\RR\), which has limit points in \(\CC\), the identity holds on \(\CC\). For the addition law, let \(F(z, w) \coloneqq e^{z+w}-e^z e^w\). For fixed real \(w\), \(z\mapsto F(z,w)\) is entire and vanishes on \(\RR\), so it vanishes on \(\CC\). Then for fixed \(z\in\CC\), \(w\mapsto F(z,w)\) is entire and vanishes on \(\RR\), so it vanishes on \(\CC\).
Exercises
A holomorphic function with a vanishing Taylor coefficient at every point is a polynomial (Tag E-IYBZP)
If \(f\) is holomorphic on a region \(\Omega\) and for each \(z_0\in\Omega\) at least one coefficient in the power series \(f(z)=\sum_{n=0}^\infty c_n(z-z_0)^n\) is zero, show that \(f\) is a polynomial.
Solution
For each \(n\ge0\), let \[ Z_n=\{z\in\Omega:f^{(n)}(z)=0\}. \] Each \(Z_n\) is closed in \(\Omega\). The hypothesis says that for every \(z_0\in\Omega\) at least one Taylor coefficient at \(z_0\) vanishes, i.e. \(f^{(n)}(z_0)=0\) for some \(n\). Hence \[ \Omega=\bigcup_{n=0}^\infty Z_n. \]
Choose a closed disk \(\overline D\Subset\Omega\). Since \(\overline D\) is a complete metric space, Baire’s theorem implies that one of the closed sets \(Z_n\cap\overline D\) has nonempty interior in \(\overline D\). Thus \(f^{(n)}\) vanishes on a nonempty open subset of \(\Omega\). By the identity theorem, \[ f^{(n)}\equiv0\quad\text{on }\Omega. \] Therefore \(f\) is a polynomial of degree at most \(n-1\) (with the case \(n=0\) giving \(f\equiv0\)).
A holomorphic function with a vanishing Taylor coefficient at every point is a polynomial (Tag E-G4N4D)
Let \(\Omega \subseteq \mathbb{C}\) be a connected open region, and let \(f: \Omega \to \mathbb{C}\) be a holomorphic function. Suppose that for every point \(z_0 \in \Omega\), at least one coefficient in the Taylor series expansion: \[f(z) = \sum_{n=0}^\infty c_n(z_0) (z - z_0)^n\] is zero (i.e. for every \(z_0 \in \Omega\), there exists some \(n \in \mathbb{N}\) such that \(f^{(n)}(z_0) = 0\)). Prove that \(f\) is a polynomial.
Solution
For \(n\ge0\), set \[ E_n=\{z\in\Omega:f^{(n)}(z)=0\}. \] Each \(E_n\) is closed in \(\Omega\) because \(f^{(n)}\) is continuous, and the hypothesis says \[ \Omega=\bigcup_{n=0}^{\infty}E_n. \]
<1>1. The region \(\Omega\) is a locally compact Hausdorff space, hence a Baire space. Therefore a countable union of closed subsets with empty interior cannot equal \(\Omega\). Thus some \(E_N\) has nonempty interior.
<1>2. Hence \(f^{(N)}\) vanishes on a nonempty open subset of \(\Omega\). Since \(f^{(N)}\) is holomorphic and \(\Omega\) is connected, the identity theorem gives \[ f^{(N)}\equiv0\quad\text{on }\Omega. \]
<1>3. If \(N=0\), then \(f\equiv0\), which is a polynomial. If \(N\ge1\), then \(f^{(N)}\equiv0\), so the Taylor expansion of \(f\) at any point terminates after degree at most \(N-1\). Equivalently, repeated integration gives a polynomial \(P\) of degree at most \(N-1\) such that \(f=P\) on \(\Omega\).
Therefore \(f\) is a polynomial.
An entire function with a vanishing Taylor coefficient at every point is a polynomial (Tag E-5P24A)
Let \(f\) be entire and for every \(a\in \CC\), and write the Laurent expansion about \(a\) as \(f(z) = \sum_{k\geq 0} c_{k, a} (z-a)^k\). Suppose that for every \(a\), there exists some \(k\) such that \(c_{k, a} = 0\). Show that \(f\) is a polynomial.
Solution
Claim: \(f^{(n)} = 0\) for some \(n\).
Let \(Z_n \da \ts{a\in \CC \st c_{n, a} = 0}\) and \(Z = \Union_n Z_n\). Since \(c_{n,a}=f^{(n)}(a)/n!\), we have \(f^{(n)}(z) = 0\) on \(Z_n\).
By hypothesis, \(Z = \CC\), so not every \(Z_n\) can be countable. Pick any \(Z_n\) which is uncountable.
Write \(Z_n = \Union_{m} Z_{m, n}\) where \(Z_{m, n} \da \DD_m(0) \intersect Z_n\) as a union of increasing discs, then not every \(Z_{m, n}\) can be finite, so pick an infinite one.
Now the infinite set \(Z_{m, n}\) has a limit point in \(\CC\), since it lies in the compact closed disk \(\overline{\DD_m(0)}\), and \(f^{(n)}(z) = 0\) on \(Z_{m, n}\), so by the identity principle \(f^{(n)}(z) = 0\) on \(\CC\).
Since \(f^{(n)}\equiv0\), all Taylor coefficients of \(f\) at \(0\) of index at least \(n\) vanish, so \(f\) is a polynomial of degree less than \(n\).