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I want to prove that there is no holomorphic function $f(z)$ on the open unit disk that satisfies: $$ f(\frac{1}{n}) = 2^{-n} \quad \quad \forall n\in\{2,3,4,\cdots\} $$

I recall that if a holomorphic function accumulates zeros, then that function is equal to zero, but I am having trouble using that fact to show that no function satisfies the above statement.

Thank you!

Gabriel Myers
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  • You're not accumulating zeroes here, so I don't think that approach is helpful. If there is such a holomorphic function $f$, what can you say about $f(0)$? $f'(0)$? – Ted Shifrin Dec 07 '22 at 20:04
  • @TedShifrin $f(0)=0=f'(0)$ can be checked, but since we have no further information on $f'$ I'm not sure how one proceeds there. I'm guessing you want to use the identity theorem? – FShrike Dec 07 '22 at 20:18
  • @FShrike No. Let's let the OP engage before we solve it here. – Ted Shifrin Dec 07 '22 at 20:37
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    https://math.stackexchange.com/q/1185504/42969, https://math.stackexchange.com/q/2904053/42969, https://math.stackexchange.com/q/241042/42969 – all found with Approach0 – Martin R Dec 07 '22 at 20:46
  • @ Martin R Unfortunately the answers in the link you cited are incorrect. I think a correct solution could be constructed by looking at properties of $ f(z/2)- f^2(z)$ – MathFont Dec 07 '22 at 21:12
  • If you know that holomorphic functions are analytic then assume that $f(z)= z^n g(z)$ with $g$ analytic and $g(0)\ne 0$, you'll get an easy contradiction. – reuns Dec 07 '22 at 21:54
  • @MathWonk: I did not check all answers, but this one looks quite correct to me: https://math.stackexchange.com/a/241498/42969 – Martin R Dec 08 '22 at 02:51
  • @ Martin R Thanks for the clarification. (Actually I only checked your first link, and that one looked wrong. But some of the other links you cited did indeed include correct solutions. ) – MathFont Dec 08 '22 at 16:44
  • Sorry, having a bit of trouble understanding the links provided (and the assumption that we can write $f(z) = z^{n} * g(z)$). Why can we write $f(z) = z^{n} * g(z)$? I am having trouble seeing why we have that $f(0) = 0$ from the assumption. After that, all of the following steps make sense to me – Gabriel Myers Dec 09 '22 at 17:25

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A variation on some of the other solutions:

Lemma. The analytic function $g(z)= f(z/2) -f^2(z)$ satisfies $g (\frac{1}{n})=0$ for every positive integer $n$.

Proof. At $z=\frac{1}{n}$, $ f(z/2)= f(\frac{1}{2n}) = 2^{-2n} = (2^{-n})^2 = f(z)^2$.

From the Lemma and the Identity Theorem, it follows that $g(z)$ is the constant $0$ at each $z$ in the unit disk. That is, $ f(z/2)= f^2(z)$ there.

This functional relation is too good to be true! Playing the right side against the left we find that the radius $R$ of convergence of the Taylor expansion of $f$ must satisfy $2R= R$ so $R= \infty$. (Thus $f$ is entire.) Since $f(0)=0$ there is a maximal open disk $\Delta$ centered at $z=0$ on which $|f|<1$. Once again the functional equation implies $\Delta =2\Delta$ so $\Delta=\infty$. Thus $f$ is a bounded entire function. It must be 0.

MathFont
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