I am new in vector and Matrix Calculus. I was studying different norm and dual norm of vectors. I found $L_1$ norm as defined as the sum of absolute value of the points in the vector. The dual of that norm is $L_\infty$ norm. I am having hard time to understand that. Particularly, from definition the dual of a norm is the supremum of $x^Ty$ where $||x||\leq1$. So $x$ is normal to the $y$?. If we plot $L_1$ norm than the dual points should be the perpendicular points. But how this produces $L_\infty$ norm?
Thanks
You did not understand. Consider the case when the vector space is $E=\mathbb{R}^2$ with the $l_1$ norm $||X||_1=|x_1|+|x_2|$. We consider the dual $E^*=L(E,\mathbb{R})$ and we show that the associated norm $||.||$ is $l_{\infty}$; if $f\in E^*$, then $f(X)=[a_1,a_2]X=a_1x_1+a_2x_2$. Assume that $\sup(|a_1|,|a_2|)=|a_1|$. Thus $||f||=\sup_{||X||_1=1}|f(X)|=\sup_{|x_1|+|x_2|=1}|a_1x_1+a_2x_2|$. We choose $sign(x_1)=sign(a_1),sign(x_2)=sign(a_2)$; then $|f(X)|=|a_1||x_1|+|a_2|(1-|x_1|)$ and the $\sup$ is obtained for $|x_1|=1$; finally , we obtain the asked result $||f||=|a_1|=||f||_{\infty}$.
Note that $E$ is not euclidean; then, there is no canonical isomorphism between $E$ and $E^*$.