(Exercise 1.1.15 in Real Analysis 2 by Tao) Let
$$X : = \{(a_n)_{n=0}^\infty : \sum_{n=0}^\infty |a_n| < \infty\}$$
be the space of absolutely convergent sequences. Define the $l^1$ and $l^\infty$ metrics on this space by
$$d_{l^1} ((a_n)_{n=0}^\infty, (b_n)_{n=0}^\infty) : = \sum_{n=0}^\infty |a_n – b_n|;$$
$$d_{l^\infty} ((a_n)_{n=0}^\infty, (b_n)_{n=0}^\infty) : = \sup_{n \in \mathbb{N}} |a_n – b_n|.$$
Show that there exist sequences $x^{(1)}, x^{(2)}, …$ of elements of $X$ (i.e. sequence of sequences) which are convergent with respect to the $d_{l^\infty}$ metric but not with respect to the $d_{l^1}$ metric. Conversely, show that any sequence which converges in the $d_{l^1}$ metric automatically converges in the $d_{l^\infty}$ metric.
Let $x^{(k)}$ be a sequence in $X$ such that $\lim_{k\to\infty}d_{l^1} ( (x^{(k)}_n)_{n=1}^\infty, x) =0$. Then, $|x_i^{(k)} – x_i| < \epsilon$ for all $i$. Thus, $x^{(k)}$ converges to $x$ in $d_{l^\infty}$ as well. This prove the second part of the question. I have difficulty in proving the first part of the question. Can you give some help?
Best Answer
As in the comments, we can take $$x^{(n)}=(\frac{1}{n},\frac{1}{n},\dots,\frac{1}{n},0,0,\dots)$$where the $n+1,n+2,...$-th entries of $x^{(n)}$ are all $0$. Then it needs to be checked that $x^{(n)}$ cannot converge in $d_{l_1}$ norm.
Edit.
Suppose that $x^{(n)}$ converges to some $x$ in $d_{l^1}$, then it can be checked that $x=(0,0,\dots)$ as follows:
Let $x(k)=$ the $k$-th coordinate of $x$. Then $$|x(k)-x^n(k)|\le d_{l^1}(x,x^{(n)})$$Taking $n\to\infty$ to see $x(k)=0$, hence $x=(0,0,\dots)$.