This is a short post about the dual p-norm, which is essentially defined by an optimisation problem. Recall that the p-norm of a vector $x\in{\rm I\!R}$ is defined as
$$\|x\|_p = \left(\sum_{i=1}^{n}|x_i|^p\right)^{\tfrac{1}{p}}.\tag{1}$$
We will show that the dual of $\|{}\cdot{}\|_p$ with $\|{}\cdot{}\|_q$, where $\tfrac{1}{p} + \tfrac{1}{q} = 1$. To do so, instead of using Holder's inequality, we will we use the KKT conditions of the optimisation problem that defines the dual norm.
The dual norm is defined as
$$\|z\|_{p,*} = \max_{\|y\|_{p} = 1} y^\intercal z = -\min_{\|y\|_{p}^{p} = 1} -y^\intercal z = -\min_{\|y\|_{p}^{p} = 1} y^\intercal z.\tag{2}$$
The dual norm is, therefore, defined via an equality-constrained optimisation problem. The Lagrangian function is
$$L(y, \lambda) = y^\intercal z + \lambda(\|y\|_{p}^{p} - 1),\tag{3}$$
and
$$\begin{aligned}\nabla_y L(y, \lambda) {}={}& z + \lambda (p|y_i|^{p-1} \mathrm{sign}(y_i) )_i, \\ \tfrac{{\rm d}}{{\rm d} \lambda}L(y, \lambda) {}={}& \|y\|_{p}^{p} - 1.\tag{4} \end{aligned}$$
We are looking for a KKT point, i.e., a point $(y^\star, \lambda^\star)$ such that $\nabla_y L(y^\star, \lambda^\star) = 0$ and $\nabla_y L(y^\star, \lambda^\star) = 0$, that is
$$\begin{aligned}z_i + \lambda^\star p|y^\star_i|^{p-1} \mathrm{sign}(y^\star_i) ={}& 0, \\ \|y\|_{p}^{p} ={}& 1.\tag{5} \end{aligned}$$
A first observation is that the only way for $\lambda^\star$ to vanish is if $z = 0$, so assuming $z\neq 0$, we have that $\lambda^\star \neq 0$. That said, if $z_i = 0$, then $y_i^\star = 0$. For those indices $i$ such that $z_i \neq 0$ we have
$$\frac{z_i}{\mathrm{sign}(y^\star_i) |y^\star_i|^{p-1}} = -\lambda^\star p,\tag{6}$$
which means that
$$\frac{z_1}{\mathrm{sign}(y^\star_1) |y^\star_1|^{p-1}} = \ldots = \frac{z_n}{\mathrm{sign}(y^\star_n) |y^\star_n|^{p-1}}.\tag{7}$$
Applying an absolute value to all terms and raising to the power $\tfrac{p}{p-1}$ we have
$$\frac{|z_1|^{\tfrac{p}{p-1}}}{|y^\star_1|^{p}} = \ldots = \frac{|z_n|^{\tfrac{p}{p-1}}}{|y^\star_n|^{p}}.\tag{8}$$
At the same time we have that $\|y^\star\|_p^p = 1$, or, what is that same
$$\sum_{i=1}^{n}|y_i^\star|^p = 1.\tag{9}$$
For convenience let us define $q=\tfrac{p}{p-1}$. By combining Equations (8) and (9) we have the linear system
$$\begin{bmatrix} |z_2|^q & -|z_1|^q \\ & \phantom{-}|z_3|^q & -|z_2|^q \\[1em] & & \ddots \\[0.7em] &&& |z_{n-1}|^q & -|z_n|^q \\ 1 & 1 & \ldots & 1 & 1 \end{bmatrix} \begin{bmatrix} |y_1^\star|^p \\ |y_2^\star|^p \\ \vdots \\ |y_n^\star|^p \end{bmatrix} {}={} \begin{bmatrix} 0\\0\\ \vdots \\ 1 \end{bmatrix}.\tag{10}$$
We can now solve this linear system to find that
$$|y_i^\star|^p = \frac{|z_i|^q}{\sum_{i=1}^{n}|z_i|^q}.\tag{11}$$
Detail: Note that if $z_i=0$ for some indices $i$, then the system of Equation (10) becomes underdetermined, but the $|y_i^\star|^p$ in Equation (11) are still a solution. From Equation (11), we have that
$$y_i^\star = \pm \left(\frac{|z_i|^q}{\sum_{i=1}^{n}|z_i|^q}\right)^{1/p},$$
where the $\pm$ means that for the time being we are not sure about the sign of $y_i^\star$. If we plug this into Equation (2) we have that
$$\|z\|_{p,*} = -\sum_{i=1}^{n}\pm \left(\frac{|z_i|^q}{\sum_{i=1}^{n}|z_i|^q}\right)^{1/p} z_i.$$
In order for $y^\star$ to be a solution of the minimisation problem in Equation (2), the sign of $y_i$ must be the same as the sign of $z_i$, in which case
$$\|z\|_{p,*} = \sum_{i=1}^{n} \left(\frac{|z_i|^q}{\sum_{i=1}^{n}|z_i|^q}\right)^{1/p} z_i = \frac{\sum_{i=1}^{n} |z_i|^{1 + \tfrac{p}{q}}}{\left(\sum_{i=1}^{n}|z_i|^q\right)^{\tfrac{1}{p}}},$$
and since $1 + \tfrac{p}{q} = q$, we have
$$\|z\|_{p,*} = \frac{\sum_{i=1}^{n} |z_i|^{q}}{\left(\sum_{i=1}^{n}|z_i|^q\right)^{\tfrac{1}{p}}} = \left(\sum_{i=1}^{n}|z_i|^q\right)^{1-\tfrac{1}{p}} = \left(\sum_{i=1}^{n}|z_i|^q\right)^{\tfrac{1}{q}} = \|z\|_{q}.$$