Holomorphic functions which converge pointwise but not almost uniformly

complex-analysispointwise-convergenceuniform-convergence

I'm looking for an example of a sequence of functions $f_n$ which are holomorphic in the open unit disk $D(0,1)$ and converge pointwise for every $z \in D(0,1)$, but are not almost uniformly convergent on $D(0,1)$ (i.e., do not converge uniformly on compact subsets). Note that such a family cannot be bounded, as then it is easy to prove that almost uniform convergence holds.

Best Answer

This is a non-trivial problem as for example there is a theorem of Osgood that says that in the conditions above, namely, $f_n$ holomorphic converges pointwise on some domain $D$, then there is a dense open subset (could be disconnected though) $D'$ of $D$ on which $f_n$ converges compactly (hence to a holomorphic function) in every connected component of $D'$.

The simplest example I know that satisfies the OP requirements is due to Newman (1976) who constructs an entire function $F(z)$ for which $F(0) \ne 0, F(re^{i\theta}) \to 0, r \to \infty, \theta$ fixed arbitrary (1), since then obviously $f_n(z)=F(nz)$ satisfies $f_n(0) \to F(0) \ne 0$ but $f_n(z) \to 0, z \ne 0$ so in particular the convergence cannot be uniform on any compact set containing zero

The example of Newman is as follows:

$g(z)=\int_0^{\infty}\frac{e^{zt}}{t^t}dt$ and then if $G(z)=g(z+4i)$ say, $G$ is entire, nonconstant and is bounded on any fixed line passing through the origin, so letting $k \ge 1$ the order of the zero of $G-G(0)$ at the origin, obviously $F(z)=\frac{G(z)-G(0)}{z^k}$ will satisfy (1) above

Edit 2024 - since it came out in another question which was slightly different but closed as a duplicate of this, let's note that the limit function can be chosen also entire with a small modification of the above, namely we now take $H$ as above st $H(0)=0$ as for example $H(z)=\frac{(G(z)-G(0))^2}{z}$ will do. Then $H(re^{i\theta}) \to 0, r \to \infty$ for a fixed $\theta$ so $H_n(z)=H(nz) \to 0$ for all $z$!

However $H_n$ cannot converge uniformly around $0$ since otherwise we would have $|H_n(z)| < C$ for all $n$ and $|z|< \delta$ but this means $H$ would be bounded since every $w$ in the plane is $nz$ for some large $n$ and $|z|<\delta$

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