We work in the space $\mathbb C^n$ and denote $v$ with $(v_1, \ldots, v_n)$
We also have $p,q > 1$ satisfying $p+q = pq$
Let $v_i = c_i|v_i|$ with every number complex, $|c_i| = 1$
Let $$w_i = \frac{\overline{c_i} |v_i|^{p/q}}{(\sum_j |v_j|^p)^{1/q})}$$
Show that $\Vert w \Vert_q = 1$ ($q$-norm) and $\sum v_iw_i = \Vert v \Vert_p$
I tried to apply the definitions and just calculate it, but couldn't find the result. I think I'm overlooking something involving the complex conjugate.
We have
$$\|w\|_q = \|(w_1, \ldots, w_n)\|_q = \frac{1}{\left(\sum_j|v_j|^p\right)^\frac1q}\left(\sum_i\underbrace{|c_i|^q}_{=1}|v_i|^p\right)^\frac1q = 1$$
Note that $\frac1q = \frac{p-1}p$ so
$$\|v\|_p^{p-1} = \left(\sum_j|v_j|^p\right)^\frac{p-1}p = \left(\sum_j|v_j|^p\right)^\frac1q$$
and $1+ \frac{p}q = p$ so
$$\sum_i v_iw_i = \frac{1}{\left(\sum_j|v_j|^p\right)^\frac1q}\sum_i \overline{c_i}v_i |v_i|^\frac{p}q = \frac{1}{\|v\|_p^{p-1}}\sum_i |v_i|^{1+\frac{p}q} = \frac{1}{\|v\|_p^{p-1}}\sum_i |v_i|^{p} = \frac{\|v\|_p^p}{\|v\|_p^{p-1}} = \|v\|_p$$
using that $\overline{c_i}v_i = |v_i|$, which follows from $v_i = c_i|v_i|$ by multiplication with $\overline{c_i}$.