## New answers tagged norms

0

There is duality condition for Lagrangian Multipliers. That relaxes the things. If additionally Slater's condition is met (one needs to check), then the duality becomes strong duality where the solution of the dual problem and the primary problem perfectly matches. If however this condition is not met, there is positive gap, but of course a much simpler ...

6

This is a comment rather than an answer but I am not entitled. The missing requirement on the two dimensional norm for the statement to be valid is that it be increasing in the natural sense that if $|x|\leq|x_1|, |y|\leq|y_1|$, then $n(x,y)\leq n(x_1,y_1)$. Norms without this property can easily be obtained by rotating a non-circular ellipse in standard ...

10

This is only an illustration to Christian Remling's beautiful answer; here are the concentric "balls" around the origin for this "norm":

10

Let $X=Y=\mathbb R$ with the absolute value norm and define $n(a,b)=\sqrt{2a^2+2b^2-3ab}$. This is a norm on $\mathbb R^2$ because it is the quadratic form of the positive definite matrix $A=\left( \begin{smallmatrix} 2 & -3/2 \\ -3/2 & 2 \end{smallmatrix}\right)$.
Then $N(v) = n(|v_1|,|v_2|)$ is not a norm because the triangle inequality fails for ...

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