Let \(f\) and \(g\) be analytic on \(U\subset \mathbb{C}\). If for some point \(x_0\in U\) and all \(k\in \mathbb{N}\)

\begin{equation*} f^{(k)}(x_0)=g^{(k)}(x_0), \end{equation*}

then \(f\equiv g\) on \(U\). [1, Corollary 1.2.5]

Proof

Let \(p\) be the power series of \(f\) at \(x_0\). Due to (0x67270472) we have \(f\equiv g\) on the domain of convergence of \(p\). Now consider

\begin{equation*} V=U\cap \{x\mid f^{(k)}(x)=g^{(k)}(x) \forall k\in \mathbb{N}\}. \end{equation*}

Our first argument implies \(V\) is open since it can be written as a union of open sets. It is also closed since the inverse image of \(0\) under \(f^{(k)}-g^{(k)}\) is closed and V is an intersection of such closed sets.

See also Link to heading

References Link to heading

  1. S. Krantz and H. Parks, A Primer of Real Analytic Functions. Boston, MA: Birkhäuser Boston, 2002. doi:10.1007/978-0-8176-8134-0