Inequality $\vert x^{*}y\vert\le \Vert x\Vert_1\Vert y\Vert_{\infty}.$

Question

I would like to prove that the following exercise :

For all $x,y\in \Bbb{K}^n$ we have $$\vert x^{*}y\vert\le \Vert x\Vert_1\Vert y\Vert_{\infty}.$$

Where $\Bbb{K}=\Bbb{R}$ or $\Bbb{C}$ and $\|{\textbf{x}}\|_{\infty}=\max\left(|x_1|, \dots, |x_n|\right)$, $\|{\textbf{x}}\|_1 = |x_1| +\ldots+|x_n|$.

Thanks to Ted Shifrin (in chat) I can use Holder inequality for $p=1$ and $q=\infty$,

So I have $$\vert \sum_{k=1}^n x^{*}_ky_k\vert\le \Vert x\Vert_1\Vert y\Vert_{\infty}$$

What is the link between $\vert x^{*}y\vert$ and $\vert \sum_{k=1}^n x^{*}_ky_k\vert$?

Answer 1

By definition of matrix product $$ x^*y=\sum_{k=1}^n x_k^* y_k $$ where $x^*$ is the conjugate transpose (and the conjugate for numbers).

Then $$ \lvert x^* y\rvert=\biggl|\,\sum_{k=1}^n x_k^* y_k\biggr|\le \sum_{k=1}^n \lvert x_k^* y_k\rvert= \sum_{k=1}^n \lvert x_k^*\rvert\,\lvert y_k\rvert= \sum_{k=1}^n \lvert x_k\rvert\,\lvert y_k\rvert $$ Now $\lvert y_k\rvert\le\|y\|_\infty$ by definition, so $$ \lvert x^* y\rvert\le\biggl(\,\sum_{k=1}^n \lvert x_k\rvert\biggr)\|y\|_\infty $$ No need of the Hölder inequality.

Answer 2

$x$ is an $n \times 1$ column vector. $x^*$ is a $1 \times n$ row vector, the conjugate transpose of $x$. When you multiply the $1 \times n$ matrix $x^*$ by the $n \times 1$ matrix $y$, you see that \begin{equation} x^* y = \sum_{k=1}^n \bar{x}_k y_k. \end{equation} (Here $\bar{x}_k$ is the conjugate of $x_k$.)