# Pathological product space norm

Let $X$ and $Y$ be two normed vector spaces and $n(\cdot, \cdot)$ be any norm on $\mathbb{R}^2$. Is it always possible to define a norm on the product vector space $X \times Y$ as $||(x, y)||_{X \times Y} = n(||x||_X, ||y||_Y)$?

Background information: the book "Advanced Calculus" by Sternberg and Loomis says the answer is negative - you have to impose an additional requirement on $n(\cdot, \cdot)$ for that to hold. However, no example is given in the text nor the exercises. Can you give me an example of such a pathological norm $n$ that fails to induce a honest normed vector space on $X \times Y$.

• Out of curiosity: what is the additional requirement? Is it also necessary? Nov 9, 2014 at 1:29
• Exactly what @fair said, it needs to be "increasing" as he defined it.
– Zuza
Nov 9, 2014 at 13:44

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 $u=(1,1)$, $v=(1,-1)$: we have that $N(u)=N(v)=1$, but $N(u+v)=2\sqrt{2}$.
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 position (as in the above answer).
• I believe some statements equivalent to this property (assuming that $||(x,y)||=n(x,y)$ is a norm on $\mathbb R^2$) are a) that $||(x,y)||_{\rm abs} = n(|x|,|y|)$ is a norm on $\mathbb R^2$, b) that $n(x,y) \ge n(x,0), n(0,y)$ for all $x,y\ge0$, or c) that the balls $B_{\rm abs}(r)=\{(x,y)\in\mathbb R^2:n(|x|,|y|)\le r\}$ are convex. Nov 11, 2014 at 14:41