# Is there a sequence of real polynomials which converge uniformly on an interval in $\mathbb{R}$ but not on a rectangle in $\mathbb{C}$?

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.

• 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. – Thomas Andrews Jul 20 '17 at 16:53
• @ThomasAndrews Yes, the title is wrong as the OP reduces to a compact interval in the body of the question. – zhw. Jul 20 '17 at 16:58
• You might want to check out math.stackexchange.com/questions/2364270/… – zhw. Jul 20 '17 at 16:59
• @ThomasAndrews: I edited the title. – Nate Eldredge Apr 28 '19 at 20:27
• @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 $\Bbb{R}$ – reuns Apr 28 '19 at 21:44

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 $$[0,1) \times [-\epsilon, \epsilon]$$.