Linear Functional is Discontinuous

Question

I'm trying to show that the linear functional $f:(\ell^1, \|\cdot\|_\infty) \longrightarrow (\mathbb{R}, \|\cdot\|)$ given by $$f((x_n)_{n \in \mathbb{N}})=\sum\limits_{n=1}^\infty x_n$$ is discontinuous. Here, $\|(x_n)_{n \in \mathbb{N}}\|_\infty=\sup\limits_{n \in \mathbb{N}} |x_n|$ and $\|\cdot\|$ is any norm on $\mathbb{R}$. I've tried to show that it is discontinuous at the origin, by exhibiting a sequence $(x^k_n)_{k \in \mathbb{N}}$ on $\ell^1$ such that $(x^k_n)_{n \in \mathbb{N}} \to (0,0,...)$ on $\ell^1$, as $k \to \infty$, but $f((x^k_n)_{n \in \mathbb{N}})\not\to 0$, but no success.

Answer 1

Equivalently, you can show that there is no constant $C > 0$ such that $$\lvert f(x) \rvert \le C \| x \|_\infty$$ for all $x \in \ell^1$. This would show that $f$ is not bounded, and hence not continuous. Consider $$x^{(k)} = (\underbrace{1,1,\ldots,1}_{k},0,0,\ldots).$$ Then each of $x^{(k)}$ is in $\ell^1$ and $\|x^{(k)}\|_\infty = 1$. However, $$\lvert f(x^{(k)}) \rvert =\sum^\infty_{n=1} x^{(k)}_n = k.$$

Answer 2

Hint: consider a sequence consisting of $n$ terms of $1/n$ and the rest $0$.