# Is there a holomorphic function on the unit disk that satisfies $f(\frac{1}{n})=\frac{1}{n^{3/2}}$?

Is there a holomorphic function on the unit disk that satisfies $f(\frac{1}{n})=\frac{1}{n^{3/2}}$?

I know that I should use identity theorem. How do I apply it here?

• Is there such a continuous function on the unit disk? How should it be defined at $0$? – Crostul Nov 13 '16 at 23:13

Assume that $f$ is holomorphic with the indicated property; we'll derive a contradiction.

Since $f$ is continuous, $f(0)=0.$ It follows that $f(z)^2=z^3$ for all $z$ in the set $\{\frac1{n}\mid n\in \mathbb{Z}^+\}\cup \{0\}.$ This set has an accumulation point, so, by analytic continuation, $f(z)^2=z^3$ for all $z$ in the unit disk.

Now define a function $g\colon\mathbb{R}\to\mathbb{C}$ by setting $g(\theta)=e^{-3i\theta/2}\,f(e^{i\theta}).$

We have $g(\theta)^2 = e^{-3i\theta}\,e^{3i\theta}=1,$ so $g(\theta)=\pm 1$ for all $\theta\in\mathbb{R}.$

The function $g$ is continuous, and the range of a continuous function on a connected set must be connected, so either $g(\theta)=1$ for all $\theta\in\mathbb{R}$ or $g(\theta)=-1$ for all $\theta\in\mathbb{R}.$

But $g(0)=f(1)$ and $g(2\pi)= e^{-3i\pi}f(e^{2\pi i})=-f(1),$ which gives us our contradiction.

it agrees with $$g(z) = z^{3/2}$$ on the set $1/n.$

$g$ cannot be made holomorphic at zero but $f$ must agree with it. So the answer is no.

it's not as simple as applying the identity theorem since $g$ is not holomorpic at zero.

However, note that $f'(0)= f(0)=0$ if $f$ is holomorphic by considering limit of finite differences. So $f$'s power expansion has first two terms zero so $|f(z)| \leq c|z|^2$ near $0$ but this is not compatible with the stated property of $f.$

So $f$ cannot be holomorphic.

• I don't follow the logic here. You can only conclude that $f$ and $g$ must agree if you have a domain where they are both holomorphic and they agree on a non-discrete set. But there is no such domain, since you cannot include $0$ in the domain if you want $g$ to be holomorphic. – Eric Wofsey Nov 13 '16 at 22:39
• Define $g(z) = z\cos (2\pi /z), z\ne 0.$ Then $g$ is holomorphic on $\mathbb D\setminus \{0\}.$ We have $g(z) = z$ on the set $\{1/n: n = 2,3,\dots \}.$ Should we conclude $z$ is not holomorphic at $0?$ – zhw. Nov 14 '16 at 2:05
• your $g$ does not vanish to order 3/2 so it does not fit into my argument as to why $f'(0)=0$ – Mark Joshi Nov 14 '16 at 2:37