# Gluing together holomorphic functions without Mergelyan theorem

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?

• You mean uniform approximation, not stronger, right? May 5, 2021 at 13:42
• yes, uniform on compacts
– Joe
May 5, 2021 at 15:35
• @Joe Why don't you think Mergelyan's theorem applies? Isn't $\Omega = \Delta \cup \Gamma$ compact, and $f$ holomorphic on the interior of $\Omega$? May 5, 2021 at 23:55
• Why do you need a proof WITHOUT Mergelian's theorem? May 6, 2021 at 2:45
• 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.
– Joe
May 6, 2021 at 7:26