Consider the unit ball $\Delta=\{|z|<1\}$ and a Lipschitz (meaning that it is the graph of some Lip. real function) segment, say $\Gamma=[1,7]$.

Consider $f$ holomorphic in a neighborhood of the ball, for convenience we take $f\in\mathcal O(\Delta_4)$ and continuous on the whole plane $\Bbb C$.

I want to approximate $f$ on $\Delta\cup\Gamma$ by entire functions.

Call $g$ the Taylor polynomial of $f$ centered in $0$ and truncated at a sufficiently high order: it is entire and approximate $f$ on $\Delta_4$ (the open ball of radius 4).

Call $h$ the entire function (provided by a theorem by P.Manne contained in his phd thesis, I cannot find online; the important part is that such a function exists) approximating $f$ on $\Gamma$.

The problem is that we don't know how does $g$ behaves on $\Gamma$ and viceversa, we don't know how does $h$ behaves on $\Delta$.

Defining a smooth function $\chi(z)$ to be 0 on $|z|<2,|z|>8$ and $1$ on $3<|z|<7$ and considering $$ \widetilde f = g+\chi\cdot(h-g) $$ this is continuous (and with compact support) on the whole plane and approximate $f$ on $\Delta\cup\Gamma$.

If allowed to use Mergelyan theorem I could approximate $\chi$ with an entire function on $|z|<9$ and conclude, but I can't.

Any hint on how to glue two entire functions without using Mergelyan thm?

  • $\begingroup$ You mean uniform approximation, not stronger, right? $\endgroup$ May 5, 2021 at 13:42
  • $\begingroup$ yes, uniform on compacts $\endgroup$
    – Joe
    May 5, 2021 at 15:35
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    $\begingroup$ @Joe Why don't you think Mergelyan's theorem applies? Isn't $\Omega = \Delta \cup \Gamma$ compact, and $f$ holomorphic on the interior of $\Omega$? $\endgroup$ May 5, 2021 at 23:55
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    $\begingroup$ Why do you need a proof WITHOUT Mergelian's theorem? $\endgroup$ May 6, 2021 at 2:45
  • $\begingroup$ Of course Mergelyan applies. I don't want to use it since I am writing a sort of generalization of it, which starts from the above explained setting. $\endgroup$
    – Joe
    May 6, 2021 at 7:26


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