bounded linear functional on $\ell^{1}$, and its relation to $\ell^{\infty}$

Question

Prove that a bounded linear functional $F$ on $\ell^1$ has representation $F(x)=\sum_{n=1}^{\infty}(c_{n}x_{n})$ where $c_{n} \in \ell^{\infty}$, and that $\|F\|_{*} = \|c_{n}\|_{\infty}$.

Answer 1

Put $c_n = F(e_n)$, then clearly, $F(x) = \sum_{n=1}^{\infty} c_n x_n$. (Hint: take $s_n = (x_1,x_2,\ldots, x_n, 0, 0, \ldots)$, then $s_n$ converges to $x$ in $l^1$; apply bounded linear property of $F$ here).

Now, because $\|e_n\|_1 \leq 1$ (hence $|c_n|\leq \|F\|_*$) for all $n$, we have $(c_n) \in l^{\infty}$ . This also implies $$ \|(c_n)\|_{\infty} \leq \|F\|_*. $$

Finally, Holder's inequality gives us: $$ |F(x)| \leq \|(c_n)\|_{\infty} \|x\|_1,$$ which implies the reversed inequality.

Answer 2

Let $c_n = F(e_n)$, where $e_n$ is the vector of zeroes except for a one in the $n$th place. Since $\|e_n\|=1$, and $F$ is bounded, we see that $|F(e_n)| =|c_n| \le \|F\|$, and so $\|c\|_\infty \le \|F\|$.

Suppose $x \in l_1$, then $x = \sum_n x_n e_n$, where $\|x\|_1 = \sum_n |x_n| < \infty$. Let $x_N = \sum_{n \le N} x_n e_n$, then it is straightforward to see that $x_N \to x$, and since $F$ is bounded we have $F(x_N) = \sum_{n \le N} c_n x_n \to F(x)$, and so $F(x) = \sum_n c_n x_n$, from which we have $\|F(x)\| \le \sum_n |c_n| |x_n| \le \|c\|_\infty \sum_n |x_n| = \|c\|_\infty \|x\|_1$, and so $\|F\| \le \|c\|_\infty$.