Assume we have:
- two vectors $u$ and $v$ such that $\Vert u \Vert > \Vert v \Vert$, with $\Vert\Vert$ being the euclidean norm
- $f$ a norm (not necessarily a norm, we only need positive definiteness and absolute homogeneity)
- $\theta$ the angle between $u$ and $v$
Proof: \begin{equation} A=f( u) -f( v) \end{equation}
let $e_{i}$ ,$e_{j}$ be basis vectors such that:
\begin{align} e_{i} \cdot e_{j} &=0\\ u&=\Vert u\Vert e_{i}\\ v&=e_{i}\Vert v\Vert \cos( \theta ) +e_{j}\Vert v\Vert \sin( \theta )\\ \end{align}
then:
\begin{align} A&=f( e_{i}\Vert u\Vert ) -f( e_{i}\Vert v\Vert \cos( \theta ) +e_{j}\Vert v\Vert \sin( \theta ))\\ &=f( e_{i})\Vert u\Vert -f( e_{i}\cos( \theta ) +e_{j}\sin( \theta ))\Vert v\Vert \end{align} thus: \begin{align} \lim _{\theta \rightarrow 0} f( e_{i})\Vert u\Vert -f( e_{i}\cos( \theta ) +e_{j}\sin( \theta ))\Vert v\Vert =f( e_{i})(\Vert u\Vert -\Vert v\Vert ) \geqslant 0 \end{align}
\begin{equation} \lim_{\theta \rightarrow 0}f(u)-f(v) \ge 0 \end{equation} So $f(u) \ge f(v)$ for a sufficiently small angle $\theta$.
I would like to know if such proof already exists so I can cite it in a paper. I'm also wondering if there are studies on the relation between the ordering given by the norm and the angle for specific norms, or for norms with additional constraints.