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I computing dual norm $ \|v\|_\ast = \sup_{\|x\|\leq1} \langle v, x\rangle $.

For example, I process like this: $x_1 = (2, 1); x_2 = (5, 10); x_3 = (8, 10)$. As norm of $x$ must be $\|x\| \leq 1$, I normalized vectors and I get this : $ x_1 = (2/\sqrt{5}, 1/\sqrt{5}); x_2 = (5/\sqrt{125}, 10/\sqrt{125}); x_3 = (8/\sqrt{164}, 10/\sqrt{164}) $.

How to choose $v$ to do the calculation? The norm of $v$ must it also be lower or equal to $1$? After calculation, I don't think I will have an interval to be able to find the supremum, hence my question of how to find the supremum? In other manuals, I also see $\|v\|_\ast = \max_{\|x\|\leq1} \langle v, x\rangle $. Which the good formula?

Martin Sleziak
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K.J Fogang Fokoa
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  • You don't choose $v$, it's given -- you're calculating $|v|_*$ so you must know it in advance. Then you consider all $x$ such that $|x|\leq 1$ and find those values of $x$ that lead to the supremum. I'm afraid I have no idea what $x_1, x_2$ and $x_3$ are supposed to be, and you really need to state what space (${\mathbb R}^2$?) you're working in as well. – postmortes Mar 02 '20 at 07:10
  • The $x$ are points I got and I work into $\mathbb{R}^2$. I normalize because the formula say $|x| \leq 1$. – K.J Fogang Fokoa Mar 02 '20 at 15:14

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I'm assuming that you're using the usual inner product on ${\mathbb R}^2$ and that your $\mathbf{x_1}, \mathbf{x_2}$ and $\mathbf{x_3}$ are all vectors that $v$ takes on in order for the question "find $\|v\|_*$" to make sense.

So, by definition of the dual norm, we have $$\|v\|_* = \sup_{\|x\|\leq 1} |v(x)| = \sup_{\|x\|\leq 1} |\langle v,x\rangle| $$ where the definition of the inner product is $$\langle v,x\rangle = v_1\cdot x_1 + v_2\cdot x_2 $$

$x_1$ and $x_2$ here without boldface are the components of $x$. Taking $v=\mathbf{x_1}$ we have that $(v_1, v_2) = (2,1)$ and so we need to find $$ \sup_{\|x\|\leq 1} |2\cdot x_1 + 1\cdot x_2| $$ By inspection we see that this is maximised when $x_1 =1$ and $x_2=0$ (note that $\|x\| = \sqrt{(x_1^2+x_2^2 )} = \sqrt{1^2+0^2}=1$) and so we obtain $$ \|v\|_* = 2 $$ You should be able to do the other two vector now by substituting in $v=\mathbf{x_2}$ and $v=\mathbf{x_3}$ and identifying the maximum.

postmortes
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