Let $(f^n,n\in\mathbb{N})$ be a sequence of complex continous functions so that $f^n(u)\longrightarrow f(u)$ uniformly to a complex continous function f if $n \longrightarrow \infty$. I addition I know that $log(f^n(u))\longrightarrow log(f(u))$ uniformly if $n \longrightarrow \infty$.
Why can we follow now also that $f^n(u)^{1/k}\longrightarrow f(u)^{1/k}$ uniformly for every k as $n \longrightarrow \infty$?
I don't know why the "log"-convergence statement is neccessary to proof that. Can someone give me a hint?
Thank you a lot!
By definition, $z^{1/k} = \exp(\log(z)/k)$. This is multi-valued in general, but presumably you're using some particular branch of log here.
Let's suppose this is the principal branch, so that $-\pi < \arg(z) = \text{Im}(\log(z)) \le \pi$, and $\arg(z^{1/k}) = \arg(z)/k$. Let $Q_j$ be the four quadrants of $\mathbb C \backslash \{0\}$, $j=1\ldots 4$, with the negative real axis included in the second quadrant but not the third. Take $N$ large enough so that for $n \ge N$, $|\log(f^n(u)) - \log(f(u))| < \pi/2$. Thus $f^n(u)$ may be in the same quadrant as $f(u)$ or a neighbouring one, except that we can't have one in $Q_2$ and the other in $Q_3$. The function $z \to z^{1/k}$ is unifomly continuous on $Q_1 \cup Q_2 \cup Q_4$ and also uniformly continuous on $Q_3 \cup Q_4 \cup Q_1$. Thus $f^n(u)^{1/k} \to f(u)^{1/k}$ uniformly on each $f^{-1}(Q_i)$, and therefore uniformly on the union of all of them.