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Suppose that $l_1 = \{ (x_n)_{n \in \mathbb{N}} | \sum|x_n| < \infty \}$ and $l_{\infty} = \{ (x_n)_{n \in \mathbb{N}}| \sup|x_n| < \infty \}$.

Show that $(l_1)^* \cong l_{\infty}$, where $(l_1)^*$ is a dual space of $l_1$.

My attempt: Define a map $L: l_{\infty} \rightarrow (l_1)^*$ given by $$L(x)(y) = \sum_{n \in \mathbb{N}}{x_ny_n}$$

where $y = (y_n)_{n \in \mathbb{N}} \in l_1$.

My aim is to show that $L$ is an isometric isomorphism.

Clearly $L$ is linear and and injective (choose $y = e_n$ for all $n \in \mathbb{N})$.

How to show that $L$ is a surjection and isometry?

Martin Sleziak
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Idonknow
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    Given a $f\in (\ell^1)^*$, consider the sequence $(f(e_1), f(e_2), \cdots )$. –  Nov 24 '15 at 07:42

1 Answers1

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Suppose $L(x) = 0$, then $L(x)y = 0$ for all $y$. Take $y=e_k$ to get $x_k = 0$, hence $x = 0$. Hence $\ker L = \{0 \}$.

Now suppose $f \in l_1^*$. Let $x_k = f(e_k)$. Since $|x_k| \le \|f\| \|e_k\| = \|f\|$, we see that $x \in l_\infty$. Then $L(x)(e_k) = f(e_k)$ and so $L(x) = f$.

Note that $|L(x)(y)| =|\sum_k x_k y_k| \le \sum_k |x_k| |y_k| \le \|x\| \sum_k |y_k| = \|x\|\|y\|$, and so $\|L(x)\| \le \|x\|$. Now choose $\epsilon>0 $ and $i$ such that $|x_i| > \|x\|-\epsilon$. Then $|L(x)(e_i)| = |x_i| \ge \|x\| -\epsilon$. Since $\epsilon>0$, we have $\|L(x)\| = \|x\|$.

copper.hat
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