What is an example of a compact convergent series of holomorphic functions, which is not uniformly convergent?

Question

If anyone can give an example as such or a definition in particular of "a compact convergent series of holomorphic functions" that would help me study a lot as I can not find one but only for a sequence.

Answer 1

The sequence of functions $f_n(z) = e^{i/n}z$ tends to $f(z)=z$ compactly but not uniformly. Visually $f_n$ rotates the input by $2 \pi/n$ around the origin. For compact convergence write $$|f(z)-f_n(z)| = |e^{i/n}z - z| = |e^{i/n}-1||z|.$$ Over the closed disc of finite radius $R$ we have $$\max\{|f(z)-f_n(z)|: |z| \le R\} \le |e^{i/n}-1|R.$$ Since $e^{i/n}\to 1$ the left-hand-side goes to zero and we have compact convergence. For uniform convergence note that for every $n$ the uniform distance.

$$\|f_n - f\|_\infty = \sup\{|f(z)-f_n(z)|: z \in \mathbb C\} =\sup\{|e^{i/n}-1|:|z|: z \in \mathbb C\}$$$$=|e^{i/n}-1|\sup\{ |z |:|z|: z \in \mathbb C\}$$

is infinite. So the left-hand-side certainly does not converge to zero!

Edit: You asked for a convergent series not a sequence. For that consider the sequence $f_n(z) = 2^{-n} z$. The series $\sum_{i=1}^\infty f_i(z)$ converges to $f(z)=z$. The proof is pretty much the same as the above. Define $g_n(z) = \sum_{i=1}^n f_i(z) = (1 - 2^{-n})z$ and show it converges compactly but not uniformly to $z$.