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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.

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  • $\begingroup$ There are no sequence of polynomials that converge uniformly on $\mathbb R$, except sequences that eventually all differ by constants. That is, if $p_i$ is the sequence of polynomials which converges uniformly on $\mathbb R$, then there is an $N$ such that for all $n>N$, $p_n(x)-p_N(x)$ is a constant. $\endgroup$ – Thomas Andrews Jul 20 '17 at 16:53
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    $\begingroup$ @ThomasAndrews Yes, the title is wrong as the OP reduces to a compact interval in the body of the question. $\endgroup$ – zhw. Jul 20 '17 at 16:58
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    $\begingroup$ You might want to check out math.stackexchange.com/questions/2364270/… $\endgroup$ – zhw. Jul 20 '17 at 16:59
  • $\begingroup$ @ThomasAndrews: I edited the title. $\endgroup$ – Nate Eldredge Apr 28 '19 at 20:27
  • $\begingroup$ @NateEldredge And to obtain a sequence converging uniformly on $\Bbb{R}$ but not on any complex open set, take something like $ f(x) = \sum_{k=1}^\infty k^{-2} \cos(2^k x), f_n(z) = \int_{-\infty}^\infty f(x) n e^{-\pi n^2 (z-x)^2}dx = \sum_{k=1}^\infty k^{-2} \cos(2^k x) e^{-\pi k^2/n^2}$, let $F_n$ be $f_n$'s $2^{2^n}$-th Taylor polynomial, then $ F_n \to f$ locally uniformly on $\Bbb{R}$ $\endgroup$ – reuns Apr 28 '19 at 21:44
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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 $[0,1) \times [-\epsilon, \epsilon]$.

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