Suppose that the \(a_n\) are pairwise distinct, \(a_n \to \infty\), and \(A_n\) are arbitrary complex numbers. Show that there exists an entire function \(f(z)\) which satisfies \(f(a_n) = A_n\).
The Fall 2006 source does not state that the interpolation nodes are distinct. That hypothesis is necessary for arbitrary prescribed values: repeated nodes with different \(A_n\) would make the interpolation conditions inconsistent.
Solution
Because the distinct sequence \(\{a_n\}\) has no finite accumulation point, the Weierstrass product theorem gives an entire function \(g\) whose zeros are exactly the \(a_n\), all simple. Thus \(g'(a_n)\ne0\) for every \(n\).
By the Mittag–Leffler theorem, there is a meromorphic function \(H\) whose principal part at \(a_n\) is \[ \frac{A_n}{g'(a_n)(z-a_n)} \] and which has no other poles. Define \[ f(z)=g(z)H(z). \] Away from the \(a_n\), this is holomorphic. Near \(a_n\) write \[ g(z)=(z-a_n)g'(a_n)+O((z-a_n)^2) \] and \[ H(z)=\frac{A_n}{g'(a_n)(z-a_n)}+O(1). \] Their product therefore has a removable singularity at \(a_n\), with \[ f(a_n)=A_n. \] After filling in these removable singularities, \(f\) is entire and interpolates all prescribed values.