Dual norm of $l_1$ of is $l_\infty$

Question

I am trying to show that $l_1$ norm's dual norm is $l_{\infty}$ norm. I have proceeded like the following:

$||z||_D = \sup \{z^Tx| ||x||_1\leq 1 \}$

Then: $ z^Tx = \sum_{i=1}^n z_i x_i \leq \sum_{i=1}^n |z_i||x_i| \leq (\max_{i=1}^n |z_i|)\sum_{i=1}^n|x_i|$

Finally since $||x||_1 \leq 1$, we have $z^Tx \leq \max_{i=1}^n |z_i|$.

With these, I am able to show that $l_{\infty}$ norm of $z$ is an upper bound of $z^Tx$ when $||x||_1\leq 1 $. But I additionally need to show that it is the least upper bound to satifsy $sup$, but I am stuck at this point.

Answer 1

We just have to pick at element of $x$ that attains it.

Given a $z$, we check it's component and look for the one with maximum norm. Say it is component $z_i$. Then we pick $x=sign(z_i)e_i$. We have $\|x\|=1$. Also,

$$z^Tx=z^Tsign(z_i)e_i=sign(z_i)z_i=|z_i|=\|z\|_\infty$$

Answer 2

It seems like you are using only finite dimensional spaces. It works for sequence spaces too. $l¹:=\{x: \mathbb N \to \mathbb C\, | \sum_{n=1}^\infty |x(n)| < \infty\}$. Take any element $x'$ from its dual space, i.e. $x': l¹ \to \mathbb C$ is linear and bounded. Its norm us defined to be \begin{align} ||x'|| = \sup_{||x||=1} |x'(x)| \end{align} Now note that the unit vectors form a Schauder basis of $l¹$. This means that for each $x\in l¹$ there exists a unique sequence $(\alpha_k)_{k\in\mathbb N}$ such that $x = \sum_{k=1}^\infty \alpha_k e_k$. We can use this to find an upper bound for $||x'||$. Take any $x\in X$ with $||x||=1$. \begin{align} |x'(x)| = \left|x'\left(\sum_{k=1}^\infty \alpha_k e_k\right)\right| = \left|\sum_{k=1}^\infty \alpha_k x'(e_k)\right| \leq \sum_{k=1}^\infty |\alpha_k| \, |x'(e_k)| \leq \sup_{k\in \mathbb N}\left(|x'(e_k)|\right) \cdot \sum_{n=1}^\infty |\alpha_k| \end{align} Now note, that the last sum must just be $||x||_1$ since $(x(k))_{k\in \mathbb N}$ is such that $x = \sum_{k=1}^\infty x(k)e_k$. (This step is much easier in finite dimensions and the Schauder basis argument is unnecessary.) Thus an upper bound for the norm is: \begin{align} ||x'|| \leq \sup_{k\in \mathbb B} |x'(e_k)| \end{align} which looks suspiciously like the $l^\infty$ norm. But we still need dot find that this is not only an upper bound. Given $\epsilon > 0$. $\exists k_0\in \mathbb N: 0 \leq \sup_{k\in\mathbb N} (|x'(e_k)|)- |x'(e_{k_0})| < \epsilon$. Consider $x_0 \in l^1$, where $x_0(k_0) = \frac{\overline{x'(e_{k_0})} }{ |x'(e_{k_0})|}$ and $x_0(k) = 0$ else. $||x_0|| = 1$. And \begin{align} ||x'(x_0)|| = |x'(e_{k_0})|> \sup_{k\in\mathbb N} |x'(e_k)| - \epsilon \end{align} This works for all $\epsilon > 0$, thus \begin{align} ||x'|| = \sup_{k\in \mathbb N}|x'(e_k)| \end{align} This is as close as we will get to the $\infty$ norm. $(l^1)'$ does not contain sequences but functionals, so there is not really a $l^\infty$ norm. But there is a canonical way to identify elements in $l^\infty$ with elements in $(l^1)'$, i.e. $(l^1)' \cong l^\infty$. If you are interested in seeing that, then write a comment.