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

Question

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 $\sup_{\Vert w \Vert_1=1} |v^\ast w|$

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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.

Answer 1

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 $|v^*w|=\|v\|_1\|w\|_\infty$.