Gauge Terrell

## Answered question

2022-07-07

Is there a real solution to: $B\le A,B\le C$? Namely, is there $s=\left(\alpha ,\beta ,\gamma ,\delta \right)\in {\mathbb{R}}^{4}$ such that $B\left(s\right)\le A\left(s\right),B\left(s\right)\le C\left(s\right)$?

### Answer & Explanation

Sanaa Hinton

Beginner2022-07-08Added 15 answers

Let $\stackrel{~}{A}\left(s\right)=A\left(s\right)-B\left(s\right)$ and $\stackrel{~}{C}\left(s\right)=C\left(s\right)-B\left(s\right)$
Since $A,B,C$ are homogeneous of degree $3$, so do $\stackrel{~}{A}$ and $\stackrel{~}{C}$. For any $s\in {\mathbb{R}}^{4}$, we have $\stackrel{~}{A}\left(-s\right)=-\stackrel{~}{A}\left(s\right)$ and $\stackrel{~}{C}\left(-s\right)=-\stackrel{~}{C}\left(s\right)$.
Using continuity of $\stackrel{~}{A}$, we can find a point $t\in {\mathbb{R}}^{4}$ with $|t|=1$ and $\stackrel{~}{A}\left(t\right)=\stackrel{~}{A}\left(-t\right)=0$. Since at least one of $\stackrel{~}{C}\left(t\right)$ or $\stackrel{~}{C}\left(-t\right)\ge 0$. We have find a $s=t$ or $-t$ such that $|s|=1$ and
$\stackrel{~}{A}\left(s\right)=0,\stackrel{~}{C}\left(s\right)\ge 0\phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}A\left(s\right)=B\left(s\right),C\left(s\right)\ge B\left(s\right)$

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