Suppose \(f\) is analytic on the open unit disk. Determine, with proof, which of the following are possible.
\(\textstyle f ( { \frac { 1 } { n } } ) = ( - 1 ) ^ { n }\) for each integer \(n > 1\)
\(\begin{array} { r } { f ( \frac { 1 } { n } ) = \exp ( - n ) } \end{array}\) for each even integer \(n > 1\) while \(\begin{array} { r } { f ( \frac { 1 } { n } ) = 0 } \end{array}\) for each odd integer \(n > 1\)
\(\begin{array} { r } { f ( \frac { 1 } { n ^ { 2 } } ) = \frac { 1 } { n } } \end{array}\) for each integer \(n > 1\)
\(\begin{array} { r } { f ( \frac { 1 } { n } ) = \frac { n - 2 } { n - 1 } } \end{array}\) for each integer \(n > 1\)
Solution
1. Part (a) is impossible.
The points \(1/n\) converge to \(0\), which lies in the open unit disk. Since an analytic function is continuous, \[ f(1/n)\longrightarrow f(0). \] But the prescribed values \[ f(1/n)=(-1)^n \] do not converge. Hence no analytic function on the disk can satisfy (a).
2. Part (b) is impossible.
For every odd integer \(n>1\), the prescription gives \[ f(1/n)=0. \] These are distinct zeros accumulating at \(0\), an interior point of the unit disk. By the identity theorem, \[ f\equiv0 \] on the disk. This contradicts the values prescribed for even \(n\), since \[ f(1/n)=e^{-n}\neq0. \]
3. Part (c) is impossible.
Since \[ \frac1{n^2}\longrightarrow0 \] and \[ f(1/n^2)=\frac1n\longrightarrow0, \] continuity gives \(f(0)=0\). Analyticity at \(0\) would then require \[ f'(0) = \lim_{z\to0}\frac{f(z)}z. \] Along the sequence \(z=1/n^2\), however, \[ \frac{f(1/n^2)}{1/n^2} = n\longrightarrow\infty. \] Thus the derivative at \(0\) cannot exist, a contradiction.
4. Part (d) is possible; one such function is\[ \boxed{ f(z)=\frac{1-2z}{1-z}. } \]
For \(\abs{z}<1\), the denominator \(1-z\) is nonzero, so this rational function is analytic on the open unit disk. For every integer \(n>1\), \[ \begin{aligned} f(1/n) &= \frac{1-2/n}{1-1/n}\\ &= \frac{n-2}{n-1}. \end{aligned} \] Hence it realizes exactly the values in (d).
5. Exactly part (d) is possible.
Steps <1>1–<1>3 rule out (a)–(c), while step 4 constructs an analytic function satisfying (d).
6. Q.E.D.
Step 5 is the requested classification.