Function on $l^{\infty}$ that assigns sum to elements of $l^1$

Question

Disprove the following statement:

$$ \text{There exists an } f \in (l^{\infty})' \text{ such that } f(x) = \sum\limits_{n = 1}^{\infty} x_n \text{ for all } x \in l^1 \subset l^{\infty} $$ I think such a function would not be continuous, but how can I prove this?

Answer 1

Hint: Let $v_n$ be the sequence in $l^\infty$ that is $1$ in the first $n$ slots and $0$ elsewhere. Note that $f(v_n)= n.$

Answer 2

Such a functional cannot be continuous. Define $a_k$ the sequence $$\left( \frac{1}{k}, \frac{1}{k+1}, \frac{1}{k+2}, \dots, \frac{1}{2k}, 0,0,0, \dots \right)$$ You have that $a_k \to 0$ in $l^{\infty}$, while $$\lim_{k \to \infty} f(a_k) = \lim_{k \to \infty} \sum_{n=0}^k \frac{1}{k+n} = \log2 \neq 0$$

Answer 3

Suppose such a function exists. Let $k \in \mathbb{N}$ be arbitrary, and let $x$ be a sequence with $x_n = 1$ for $1 \leq n \leq 2k$ and $x_n = 0$ otherwise. Then $$ ||f(x)|| = 2k > k = k \cdot 1 = k \cdot ||x||_{\infty} $$ so by definition, $f$ is not continuous, contradicting $f \in (l^{\infty})' = B(l^{\infty}, \mathbb{F})$.