$(x^{(k)})$ is a sequence of vectors, where $x^{(k)} = (x^{(k)}_1,...,x^{(k)}_n)$, and $x = (x_1,...,x_n)$.
I was told that $\|x^{(k)} - x\|_1 \le n\ max(|x^{(k)}_i - x_i|)$, but don't see how. I know that $\|x - y\|_1 \le n\|x - y\|_\infty$, but here's how I see the expansion of the norm:
$\|x^{(k)} - x\|_1 = |x^{(1)}_1 - x_1|+...+|x^{(1)}_n - x_n|+|x^{(2)}_1 - x_1|+...+|x^{(m)}_n - x_n|$
$n\max|x^{(k)}_i - x_i| = n\max (|x^{(1)}_i - x_i|,...,|x^{(m)}_i - x_i|)$
So from this expansion, I don't see how the inequality would hold true. What am I interpreting wrong?
The example I have here is different from the question I've posted, because I would just like to know, in general, the difference between the expansion of a norm $\|x^{(k)} - x\|$ and $\|x^{(k)}_i - x_i\|$
This question also arises from a proof about showing that a sequence of vectors $(x^{(k)})$ converges to $x$ iff $x^{(k)}_i$ converges to $x_i$. In my notes that proof used an inequality that $|x^{(k)}_i - x_i| \le \| x^{(k)} - x \| \le n\max (|x^{(k)}_i - x_i|)$
I'm sure you are intended to take the max over $i \in \{1, ..., n\}$, not over $k$. Notice that for each positive integer $k$ we have (by definition of the 1-norm):
\begin{align} ||x^{(k)}-x||_1 &= |x_1^{(k)}-x_1| + |x_2^{(k)}-x_2| + ... + |x_n^{(k)}-x_n| \\ &\leq n\max_{i\in\{1, ..., n\}}\{|x_i^{(k)}-x_i|\} \\ &= n||x^{(k)}-x||_{\infty} \end{align} where the last equality just uses the definition of the $\infty$-norm. Notice that $k$ is held fixed consistently throughout.