$\ell^1$ is not complete for the norm $\|\cdot\|_\infty$

Question

Let $\ell^\infty = \{ (u_n) | u_n \in \mathbb{R}$ and $\sup_{n \in \mathbb{N}}|u_n| < \infty \}$ and $\ell^1 = \{ (u_n) | u_n \in \mathbb{R}$ and $\sum_{n=1}^{\infty} |u_n| < \infty \}$

We know that $\ell^1 \subset \ell^\infty$, so $\ell^1$ inherits the norm $\|\cdot\|_\infty$ of $\ell^\infty$. We want to show that :

1. $\ell^1$ is not complete for the norm $\|\cdot\|_\infty$.

I considered the sequence $(u^{(n)})$ defined by $u^{(n)}_k = \frac{1}{k}$ if $k \leq n$ and $u^{(n)}_k = 0$ otherwise. This sequence doesn't converge to an element $l \in \mathcal{l}^1$. For if it does, we have necessarily that $l_k = \frac{1}{k}$ and $l \notin \ell^1$. Moreover, my guess is that this sequence is Cauchy. But how exactly do we prove that ?

My second question is:

2. We still endow $\ell^1$ with the norm $ \|\cdot\|_\infty$. Let $\phi : \ell^1 \times \ell^1 \to \mathbb{R}$ be such that $\phi(U,V) = \sum_{n=1}^{\infty} u_n v_n$. Show that $\phi$ is not continuous.

I don't know which sequences $U$ and $V$ I might take with the assumption that $\phi$ is continuous in order to get a contradiction.

Can someone help ?

Answer 1

For the first one, if $m>n$ $$ \|u^{(n)}-u^{(m)}\|_\infty=\sup\{(0,\ldots,0,\frac1{n+1},\frac1{n+2},\ldots,\frac1m)\}=\frac1{n+1}. $$ So the sequence is Cauchy.

For the second one, let $$u^{(n)}=v^{(n)}=\bigg(\underbrace{\frac1{\sqrt n},\ldots,\frac1{\sqrt n}}_{n \text{ times }},0,\ldots\bigg).$$ Then $\|u^{(n)}\|_\infty=\|v^{(n)}\|_\infty=\frac1{\sqrt n}\to0$, while $\phi(u^{(n)},v^{(n)})=1$ for all $n$.