Let's consider, for $p\ge1$, the $p$-norm $\|\cdot\|_p:\Bbb C^n\to[0,+\infty)$, defined by $\|x\|=\sqrt[p]{|x_1|^p+\ldots+|x_n|^p}$, where $x=(x_1,\ldots,x_n)\in\Bbb C^n$.
I know that for any $x\in\Bbb C^n$ we have that $\displaystyle\lim_{p\to\infty}\|x\|_p=\|x\|_\infty$, where $\|x\|_\infty=\max(|x_1|,\ldots,|x_n|)$. Given $x\in\Bbb C^n$, instead of $\|x\|_p$ we could consider $\|x\|_k$ for $k\in\Bbb N$, so we have a sequence $(\|x\|_k)_{k\in\Bbb N}$ that converges to $\|x\|_\infty$. This means that the sequence of functions $f_k=\|\cdot\|_k$ converges pointwise to $f=\|\cdot\|_\infty$.
My question is if that sequence is converging also uniformly. I don't think so, but I wasn't able to get any counterexample. Any ideas?
Thanks in advance.
I finally got a contradiction: Suppose $f_k$ converges uniformly to $f$ and take some $\varepsilon>0$. Then there is some $m\in\Bbb N$ such that $\big|\|x\|_k-\|x\|_\infty\big|<\epsilon\quad\forall x\in\Bbb C^n\quad\forall k\ge m$.
Consider $x=(l,\ldots,l)\in\Bbb C^n$ with $l\in\Bbb N$. Then $\big|\|x\|_k-\|x\|_\infty\big|=\big|\sqrt[k]{l^k+\ldots+l^k}-\max(l,\ldots,l)\big|=\big|\sqrt[k]n\,l-l\big|=(\sqrt[k]n-1)l$, since $\sqrt[k]n-1\ge0$ for all $k$.
Taking $k=m$ we get $\big|\|x\|_m-\|x\|_\infty\big|=(\sqrt[m]n-1)l<\varepsilon$ by hypothesis, but $(\sqrt[m]n-1)l\xrightarrow{l\to\infty}+\infty$, which is clearly a contradiction. Therefore the convergence wasn't uniform.