TITLE: Is there a sequence of real polynomials which converge uniformly on an interval in $\mathbb{R}$ but not on a rectangle in $\mathbb{C}$?
QUESTION [6 upvotes]: In particular I wondered about the following: The Weierstrass-function $\mathcal{W}$ is continuous and nowhere differentiable. By the Stone-Weierstrass-Theorem we can approximate $\mathcal{W}$ on $[0,1]$ uniformly by real polynomials. Let $p_n(x)$ be such a sequence of polynomials. Now we consider the $p_n$ as complex polynomials. On $[0,1]$ the $p_n$ of course still converge pointwise to $\mathcal{W}$. On $[0,1]\times i[-\frac{1}{2},\frac{1}{2}]\subseteq\mathbb{C}$ however this convergence can not be uniform anymore, as this would imply holomorphy on $(0,1)\times i(-\frac{1}{2},\frac{1}{2})$ which would imply real differentiability on $(0,1)$.
I find this very unintuitive, so i would like to see a concrete example of a sequence of polynomials converging uniformly on some $[a,b]\subseteq \mathbb{R}$ but not converging uniformly on any $[a,b]\times i[-\epsilon,\epsilon]\subseteq\mathbb{C}$, if possible with a direct verification that this is (not) the case.

REPLY [3 votes]: As a simple example, consider $p_n(x) = \sum_{k=1}^n \frac{(-x)^k}{k}$, which is the $n$th degree Taylor polynomial for $-\ln(1+x)$.  The power series $\sum_{k=1}^\infty \frac{(-x)^k}{k}$ converges pointwise on $[0,1)$ (ratio test) and also at $x=1$ (it is the alternating harmonic series).  So by Abel's theorem, the series converges uniformly on $[0,1]$.  However, the ratio test also makes it clear that the series $\sum_{k=1}^\infty \frac{(-z)^k}{k}$ diverges at every $z$ with $|z|>1$; in particular, it diverges at $z=1 \pm \epsilon i$ for any $\epsilon > 0$.
This shows that the sequence of polynomials $p_n$ converges uniformly on $[0,1]$, but does not even converge pointwise on any rectangle $[0,1] \times [-\epsilon, \epsilon] \subset \mathbb{C}$, or even $[0,1) \times [-\epsilon, \epsilon]$.