0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.