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Consider a compact set $K\subset \mathbb{R}$ with positive measure (i.e. $\mu(K)>0$), and for $z\in\mathbb{C}\backslash K$, define the holomorphic function $f$ on $\mathbb{C}\backslash K$ by \begin{equation} f(z):=\int_K \frac{dt}{t-z}. \end{equation}

Now the question is can $f$ be extended to an entire function? Clearly $f(z)\to 0$ as $z\to \infty$ in every direction, and I'm not familiar with entire functions of this kind, this seems to me a realy pointless question :(

IIIsomorphiii
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    Just look at simple examples like $K$ an interval and see what happens; in general there will be singularities at boundary points - not sure why you say $f(z)$ goes to zero at zero, maybe you meant at infinity which immediately should have told you that such an extension is not possible by liouville – Conrad Nov 16 '23 at 13:43
  • @Conrad Sorry I made a typo... Thank you for your answer! But I've read here that there exist entire functions that are bounded in every direction $z\to\infty$, and I'm extremely confused as in my mind this implies contradiction by Liouville immediately. – IIIsomorphiii Nov 16 '23 at 14:34
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    The functions you are talking about are bounded on lines not unrestrictedly like here; also they are of infinite order while here it is trivial to see that the function is of zero order since it is unrestrictedly bounded outside a disc including the compact; here it is not difficult to show that the function extends at any interior point of $K$ and is singular at any point that's an end of an interval in the complement of $K$ and I suspect that it will be singular at all non interior points of $K$ even if they are not end points of complement intervals since those should be limits of such – Conrad Nov 16 '23 at 15:04
  • @Conrad I seemed to have overcomplicated the problem, this really helped me a lot~ – IIIsomorphiii Nov 16 '23 at 15:52

1 Answers1

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Choose $R > 0$ such that $K$ is contained in the disk with center at the origin and radius $R$. For $|z| > 2R$ and $t \in K$ is $$ |t-z| \ge |z| - \frac 12|z| =\frac 12 |z| $$ and therefore $|f(z) | \le \frac {2}{|z|} \mu(K)$ for $|z| > 2R$.

This shows that $$ \lim_{z \to \infty} f(z) = 0 \, , $$ which is a stronger statement than being “bounded in every direction $z \to \infty$.”

If $f$ could be extended to an entire function then that function would be bounded on $\Bbb C$, and therefore constant (Liouville's theorem), and that constant is necessarily zero.

Martin R
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