Vrftsjtryk

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

$||z||_D = sup {z^Tx| ||x||_1leq 1 }$

Then:
$ z^Tx leq sum_{i=1}^n |z_ix_i| = 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||_1leq 1 $. But I additionally need to show that it is the least upper bound to satifsy $sup$, but I am stuck at this point.

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,

$$z^Tx=z^Tsign(z_i)e_i=sign(z_i)z_i=|z_i|=|z|_infty$$

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 $xin l¹$ there exists a unique sequence $(alpha_k)_{kinmathbb N}$ such that $x = sum_{k=1}^infty alpha_k e_k$. We can use this to find an upper bound for $||x'||$. Take any $xin X$ with $||x||=1$.
begin{align}
|x'(x)| = left|x'left(sum_{k=1}^infty alpha_k e_kright)right| = left|sum_{k=1}^infty alpha_k x'(e_k)right| leq sum_{k=1}^infty |alpha_k| , |x'(e_k)| leq sup_{kin 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 $(x(k))_{kin mathbb N}$ 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_{kin 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 $epsilon > 0$. $exists k_0in mathbb N: 0 leq sup_{kinmathbb N} (|x'(e_k)|)- |x'(e_{k_0})| < epsilon$. 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_{kinmathbb N} |x'(e_k)| - epsilon
end{align}
This works for all $epsilon > 0$, thus
begin{align}
||x'|| = sup_{kin 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.