TITLE: Norm of a linear map is not attained
QUESTION [6 upvotes]: Prove that the norm of the linear functional
$$\phi: l^1 \ni \{x_n \} \rightarrow \sum_{n=1} ^{\infty} (1 - \frac{1}{n} )x_n \in \mathbb{K}$$
equals one but there doesn't exist a sequence $ \{x_n \} \in \mathcal{l}^1$ such that $|| \{ x_n \} || \le 1$ and $| \phi ( \{ x_n \} ) | =1$
My problem is that I don't know how to show that the norm equals one - it's easy to show that it's less or equal $1$.
I guess I need to contruct a sequence of elements of $l^1$ whose norms tend to one.
And how to prove that there doesn't exist $\{ x_n \} \in l^1$ for which the value $1$ of the norm is attained?
Could you explain that to me, please?

REPLY [4 votes]: For the first question: Let a sequence be defined by letting $x_m=1$, and $x_k=0$ for $k \neq m$. Then what happens when $m$ goes to infinity?
For the second question:
$$
| \phi ( \{ x_n \} ) | \leq \sum_{n=1} ^{\infty} |(1 - \frac{1}{n} )x_n| < \sum_{n=1} ^{\infty} |x_n| \leq 1.
$$
The strict inequality is true for any sequence where $||\{x_n\}|| \neq 0$.

REPLY [2 votes]: Let denote $(e_n)_{n\in \mathbb{N}}$ the canonical basis. Then $\phi (e_n)= (1-1/n)$, so for every $n\in \mathbb{N}$, $\| \phi \| \geq (1-1/n)$.
For your second question, observe that $|\phi (x) |\leq \sum |(1-1/n)x_n| < \sum |x_n | =1$, if $\|x \| = 1$.

REPLY [2 votes]: Hints: To show the norm is at least $1$, try it on the "standard basis vectors" that have a single $1$ and everything else $0$.
To show there is no member of $\ell^1$ where that is attained, note that
if $x_m \ne 0$, $|\phi(x)| \le \|x\| - |x_m|/m$.