Question:
On the space $\ell^1$ for $x=(\alpha_1,\alpha_2,\ldots)\in{\ell^1}$, define $f(x)=\sum\limits_{n=1}^\infty \alpha_n$ Prove that $f$ is not continuous with respect to $\|x\|_\infty =\sup_n|\alpha_n|$.
This is my proof:
Since $f$ is a linear map, $f$ is continuous iff $f$ is bounded. Assume for the sake of contradiction that $f$ is bounded. This implies that there exist $k\gt 0$. such that $\|f(x)\|\le k\|x\|_\infty \text{ for all } x\in \ell^1$ In particular, $k$ may be $1$.
Now, let $x=(\alpha_1,\alpha_2,\ldots)\in \ell^1$ where $\alpha_i \ge 0$ for all $i$.
$\|f(x)\|=\|\sum\limits_{n=1}^{\infty}\alpha_{n}\|$ Since for each $i$ $\alpha_{i}\gt 0$,
$=\sum\limits_{n=1}^\infty |\alpha_n| \gt\sup_n|\alpha_n|=\|x\|_\infty$
This implies $\|f(x)\|\gt\|x\|_\infty$
Contradicting my first line of proof. Hence $f$ is not bounded, i.e. $f$ is not continuous.
My problem here is that I failed to believe myself, I think something is wrong in the proof. can anyone help me out?