# Showing $\sup_{\Vert w \Vert_\infty} |v^\ast w| = \Vert v \Vert_1$ for $v \neq 0$

## Problem

Show, for $$v \neq 0 \in \mathbb{C}^n$$ $$\sup_{\Vert w \Vert_\infty=1} |v^\ast w| = \Vert v \Vert_1$$

And find the similar equality for $$\sup_{\Vert w \Vert_1=1} |v^\ast w|$$

## Try

Since $$|v^\ast w| = \cos\theta \Vert v \Vert_2 \Vert w \Vert_2$$, geometrically

$$\sup_{\Vert w \Vert_2=1} |v^\ast w| = \Vert v \Vert_2$$

is direct, since this means the Euclidean distance from origin of the orthogonally projected vector on $$w$$.

However, $$\Vert w \Vert_\infty$$ and $$\Vert w \Vert_1$$ form kind of "boxes", so I'm stuck at how I should proceed.

First show that $$|v^*w|\leq\|v\|_1\|w\|_\infty$$. This is a special case of the Holder's inequality, which is particularly easy: $$\begin{split} |v^*w|&=\left|\sum_{i=1}^n \bar{v}_i w_i\right| \leq\sum_{i=1}^n |v_i||w_i| \leq\left(\sum_{i=1}^n |v_i|\right)\left(\max_{1\leq i\leq n}|w_i|\right) =\|v\|_1\|w\|_\infty. \end{split}$$
The inequality is attained if we can find a $$w$$ such that $$\bar{v}_i w_i=|v_i|$$ and $$\|w\|_\infty=1$$. This is easily done. If $$v_i=|v_i|e^{\iota \theta_i}$$ (where $$\iota$$ is the imaginary unit), then choosing $$w_i=e^{\iota\theta_i}$$ gives $$\bar{v}_iw_i=|v_i|e^{-\iota\theta_i}e^{\iota\theta_i}=|v_i|$$ and $$|w_i|=1$$ so $$\|w\|_\infty=1$$. Hence $$|v^*w|=\|v\|_1\|w\|_\infty$$.