Suppose \(f\) is analytic on \(\DD^\circ\). Determine with proof which of the following are possible:
\(f\qty{1\over n} = (-1)^n\) for each \(n>1\).
\(f\qty{1\over n} = e^{-n}\) for each even integer \(n>1\) while \(f\qty{1\over n} = 0\) for each odd integer \(n>1\).
\(f\qty{1\over n^2} = {1\over n}\) for each integer \(n>1\).
\(f\qty{1\over n} = {n-2 \over n-1}\) for each integer \(n>1\).
Solution
Part a: Not possible: if \(f\) is holomorphic then \(f\) is in particular continuous, so \[ f(0) = f(\lim 1/n) = \lim f(1/n) = \lim (-1)^n ,\] which does not converge.
Part b: Not possible: note that \(1/n\) has a limit point, so if \(f(1/n)=0\) then \(f\equiv 0\) on \(\DD\) by the identity principle. In particular, we can not have \(f(1/n) = e^{-n}>0\).
Alternatively, note that a holomorphic \(f\) must have isolated zeros, while \(z_0=0\) is forced to be a zero of \(f\) by continuity, which has infinitely many zeros of the form \(1/n\) in any neighborhood.
Part c: Not possible: suppose so, then by continuity, we have \[ f(0) = f(\lim 1/n^2)= \lim f(1/n^2)=\lim 1/n = 0 ,\] so \(z_0=0\) is a zero. Now defining \(g(z) = z^{1\over 2} \da e^{1\over 2 \log(z)}\) on \(U \da \CC\sm(-\infty, 0]\) extending this continuously to zero by \(g(0)= 0\) yields \(g(z) = f(z)\)on \(\ts{1/n^2 \st n>1}\union\ts{0}\), so \(g(z) \equiv f(z)\) on \(U\). But then \(g\equiv f\) on \(\DD\), and \(g\) is not holomorphic on all of \(\D\), contradicting that \(f\) was holomorphic on \(\DD\).
Part d: Yes: note that this forces \(f(0) = \lim {n-2\over n-1} = 1\) by continuity at \(z=0\). We can write \[ {n-2\over n-1} = {1 - 2\cdot{1\over n} \over 1 - {1\over n}} ,\] so define \(g(z) \da {1-2z\over 1-z}\). Then \(g(1/n) = f(1/n)\) for all \(n\) and \(g(0) = 1= f(0)\), so \(g=f\) on a set with an accumulation point making \(g\equiv f\) on \(\DD\). Note that \(g\) is holomorphic on \(\DD\), since it has only a simple pole at \(z_0 = 1\).