Can anyone confirm that this proof is correct? I don't see any reason for it to be incorrect, but I feel like I may be missing something.
Theorem:
There exists no function $f$ such that $f$ is analytic in $\mathbb C$ and $f(1/n) = |1/n^3|$ for any $n \in \mathbb Z\backslash \{0\}$.
Proof:
In order to prove this, we will do so by contradiction. Consider the function $f$ described above. Since $0$ is an accumulation point for $S = \{1/n\ |\ 0 < n \in \mathbb Z\backslash \{0\}\}$ and $f(s) = s^3$ for all $s \in S$, we know that $f(z) = z^3$ in the positive half of the complex plane. Since $0$ is an accumulation point for $S^- = \{1/n\ |\ 0 > n \in \mathbb Z\backslash \{0\}\}$ and $f(s) = -s^3$ for all $s \in S^-$, we know that $f(z) = -z^3$ in the negative half of the complex plane. Now, consider the point $z_0 = 0$. Since $f$ is analytic, we know that there exists $r>0$ such that $f$ can be expressed as a power series on the set $D(0,r)$. For $\Re(z) > 0$, we find that $f(z) = z^3$ in the set $D(0,r)$. For $\Re(z) < 0$, we find that $f(z) = -z^3$ in the set $D(0,r)$. Since $z^3=-z^3$ in the disk $D(0, r)$, we have reached a contradiction. Therefore, the theorem is true. $QED$