$l_1$ equipped with the sup norm is NOT a Banach Space

Question

Prove that $l_1 = \{ x = (x_k)_{k\in\mathbb{N}}\subset \mathbb{R};\ \sum_{k\in\mathbb{N}}\ |x_k| < +\infty \}$ equipped with the norm $\| x\| = \mathrm{sup}_{k\in\mathbb{N}} |x_k|$ is NOT a Banach Space.

I've tried to solve it considering the sequence $x_n = (\frac{1}{k^{1 + 1/n}}),\ \forall n \in \mathbb{N}$

Since $\sum_{k\in\mathbb{N}}\frac{1}{k^{1 + 1/n}} < +\infty$ (because $1+1/n > 1$), $(x_n)_{n\in\mathbb{N}}$ is a sequence in $l_1$. We also have that $x_n \longrightarrow (1/k)_{k\in\mathbb{N}},$ as $n\longrightarrow +\infty$, and this one is not in $l_1$.

I would like to prove that $(x_n)_{n\in\mathbb{N}}$ is a Cauchy sequence, but I don't have a clue how to do it.

Answer 1

Let me propose you an example I find easier.

So $\ell^\infty$ is a Banach space with the sup norm, and $\ell^1$ is a subspace of $\ell^\infty$. Hence what you want boils down to: $\ell^1$ is not closed in $\ell^\infty$.

For every $n\geq 1$, consider the finitely supported sequence $$ x^{(n)}=\left(1,\frac{1}{2},\frac{1}{3},\ldots,\frac{1}{n},0,\ldots\right). $$ Clearly, every $x^{(n)}$ is in $\ell^1$. It is also easy to see that $x^{(n)}$ converges to $$ x=\left(1,\frac{1}{2},\frac{1}{3},\ldots,\frac{1}{k},\frac{1}{k+1},\ldots\right) $$ in $\ell^\infty$, for the sup norm.

But $x$ is not in $\ell^1$ as the harmonic series diverges. So $\ell^1$ is not closed in $\ell^\infty$.

Answer 2

Let $x_n$ be any sequence that converges to zero, but $\sum_n |x_n| = \infty$.

Let $[y_k]_n = \begin{cases} x_n & n \le k \\ 0 & \text{otherwise}\end{cases}$. Then $y_k \in l_1$ for all $k$, and $y_k$ is a Cauchy sequence (in the $l_\infty$ norm). Since $l_\infty$ is complete, $y_k$ converges to some $y \in l_\infty$ (it is easy to see that $[y]_n = x_n$), but clearly $y \notin l_1$.