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While thinking about the identity theorem the following question came in my mind:

Let $A\subset \mathbb{C}$ be a set with an accumulation point in $\mathbb{C}$. What properties does a function $f:A\to \mathbb{C}$ need, such that we can find an extension $\tilde{f}:\mathbb{C} \to \mathbb{C}$ with $\tilde{f}(z)=f(z)$ for all $z\in A$ and $\tilde{f}$ is an entire function ?

My actual thoughts are:

  • $f$ must be local lipschitz continuous
  • We could try to calculate the coefficients of the Taylor Series, (my first thought is doing something like a limit of polnoymial interpolation, so taking a converging sequence $(b_n)_{n\in \mathbb{N}}$ with values in $A$ and solving $V_{(b_n)_{n\in \mathbb{N}}}\cdot (a_n)_{n\in \mathbb{N}}= (f(b_n))_{n\in \mathbb{N}}$, where $V$ is an $\infty \times \infty$ Vandermonde-matrix). If this work we have the only possible canditate for $\tilde{f}$ namely $$\tilde{f}(z)=\sum_{n=0}^\infty a_n \cdot z^n$$ because of the identity theorem. So we just need to check whether $f(z)=\tilde{f}(z)$ for all elements of $A$, and wheter the radius of convergence of $\tilde{f}$ is infinity. From my very naive point of view this should work iff $f$ admits such an extension, because the sums in the linear equation are unconditional convergent. So what are conditions for the equation to have a solution?
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  • $\begingroup$ Have you tried this for any particular examples? Like $\{\frac{1}{n}\}\cup\{0\}$? PS, doesn't locally Lipschitz imply locally continuous which implies continuous? $\endgroup$ – Alex Youcis Sep 3 '13 at 7:26
  • $\begingroup$ @AlexYoucis yes for sure as my extension is holomorphic every restriction must be continuous. when you know the values of $\{\frac{1}{n}\}$ you already know the value on $f(0)$ because of continuity. $\endgroup$ – Dominic Michaelis Sep 3 '13 at 8:27
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In fact one can compute potential coefficients of the power series of $\tilde{f}$. We have the accumulation point $x$ and from definition we have a sequence $(z_n)_{n\in \mathbb{N}}$ which converges to $z$ and $z_k\neq z$ for all $k\in \mathbb{N}$.

Now we know that $$f\left( \lim_{n\to \infty} z_n \right)=\lim_{n\to \infty} f(z_n)$$ must hold else there can't be an continuous extension and furthermore no holomorphic one.

Wlog we can say that $x=0$. Let $(a_n)_{n\in \mathbb{N}}$ be the coefficients of the power series representation of $\tilde{f}$.

From the given values we already can compute the coefficients. We get $a_0$ from the continuity and furthermore we know that whenever $f$ admits an holomorphic extension \begin{align*} \lim_{z\to 0} \frac{\tilde{f}(z)-\sum_{k=0}^{n-1} a_k z^k }{z^n} &= \lim_{z\to 0} \frac{\sum_{k=n}^\infty a_k z^k}{z^n}\\ & =\lim_{z\to 0} \sum_{k=n}^\infty a_k z^{k-n} \\ &= a_n \end{align*} must hold.

So taking our sequence which is a to $0$ converging sequence we can compute those coefficients starting with $a_1$, $a_2$ and so on.

For sure this is only possible when those limits exist, but furthermore the existence of the limits doesn't imply that our function admits an holomorphic extension, even if the radius of convergence of $\sum_{n=0}^\infty a_n \cdot z^n$ is infinity. Here taking $f(z)=\exp\left(-\frac{1}{z^2}\right)$ and $A=\{ \frac{1}{n}: n \in \mathbb{N}\}$ gives a counter example, as all coefficients will be zero, but obviously our function is not zero.

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