This map $f$ is not continuous with respect to sup norm 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?
 A: Boundedness means:
$$(\exists C)(\forall x) \|f(x)\|\le C\|x\|.$$
Negation of this is
$$(\forall C)(\exists x) \|f(x)\|> C\|x\|.$$
Note two important things:


*

*You only have shown $C=1$ is not working.

*You only need to show existence of one such $x$ for a given $C$. (In your attempted proof above you worked with arbitrary $x$.)


So the question is: If a constant $C$ is given, can you find a sequence $x$ such that $\|f(x)\|> C\|x\|$?

EDIT (added after seeing OP's comments):
Maybe this could clarify what's going on.
Problem: Show that the sequence $a_n=2^n$ is bounded.
A sequence is bounded if there exists a constant $C$, such that $|a_n|\le C$ for each $n$. In this case -- since $a_n$ is positive -- this is the same as $a_n \le C$.
Solution 1: Since $a_1=2>1$, the above property fails for $C=1$. So $a_n$ is not bounded.
Solution 2: We can show by induction that $2^n>n$ holds for $n=1,2,\ldots$. Thus for every given $C>1$ we can take $n=\lceil C \rceil$ and we have $C\le n < 2^n=a_n$. Thus the sequence $a_n$ is not bounded by any given constant $C$. 
Which of the above solutions is correct? Do you see the similarity with your proof of unboundedness of $f$?

So far I have tried to get you to solve the problem by yourself, however, if there is still a problem, you can have a look at this (Spoiler alert - the solution appears when you 
move your mouse bellow):

 Suppose that $C$ is a positive integer. Then we can choose a sequence $x$ such that $x=(\underset{(C+1)\text{-times}}{\underbrace{1,1,1,\dots,1}},0,0,0,\dots)$. Then $\|x\|=1$ but $f(x)=C+1$, hence $|f(x)|>C\|x\|$. 
 This works for any positive integer $C$, so the function $f$ is not bounded.

A: You can also argument as follows:
Consider the sequence $x_n\in\ell^1$ given by
$$
x_n=\left(1,\frac{1}{2^{1+\frac{1}{n}}},\frac{1}{3^{1+\frac{1}{n}}},\ldots,\frac{1}{k^{1+\frac{1}{n}}},\ldots \right)
$$
Note that for all $n\in\mathbb{N}$ we have $\|x_n\|_{\infty}=1$. In other words $x_n$ is a sequence in the unit ball. To show that $f$ is not continuous it is enough to show that $|f(x_n)|\to\infty$.  This can be done using the integral test since 
$$
n=\int_{1}^{\infty} \frac{1}{x^{1+\frac{1}{n}}} \ dx\leq \sum_{k=1}^{\infty}\frac{1}{k^{1+\frac{1}{n}}}=|f(x_n)|.
$$  
