TITLE: Can a function be analytic and satisfy $f\left(\frac 1 n\right) =\frac{1}{\log{n}}.$?
QUESTION [8 upvotes]: Let $\Omega = \{z\in\mathbb{C}:\,|z|<2\}$. Prove or disprove that there exists an analytic function $f:\Omega\rightarrow\mathbb{C}$ such that for $n\geq2$:
$$f\left(\frac 1 n\right) =\frac{1}{\log{n}}.$$
Usually this question would fall under the uniqueness theorem for analytic functions; an example where the uniqueness theorem works is in disproving the existence of an analytic $f$ satisfying $$f\left(\frac 1 n\right) =\frac{(-1)^n}{n^2}.$$
But in the case of the $\log$, I think the issue of existence is more "substantial". 
Is it true that if such $f$ were to exist, then $f(z) = -\frac{1}{\log z}$? And if so, what is the main reason this function cannot be analytic in $\Omega$?
Thanks in Advance!

REPLY [13 votes]: Assume that such a function exists. Then (by continuity) $f(0) = 0$. Thus we can write
$$
f(z) = zg(z)
$$
for some holomorphic $g$. Plug in $z=1/n$ to get
$$
g(1/n) = nf(1/n) = \frac{n}{\log n}
$$
but if we let $n\to\infty$, it follows that $g$ is unbounded near $0$, which is a contradiction.

REPLY [2 votes]: If we require $f(z)$ to be analytic then $$f(z)=-\frac{1}{\log z}\tag1$$
is the only solution, because having a function defined at an accumulation point of its domain, we can expand it to its whole domain. The accumulation point in this case is $z=0$. Function $f$ cannot be analytic at $\Omega$ as $(1)$ isn't.