Differential exclusion of a large interior weighted angle maximum

Suppose $p$ is an interior local maximum of ${\tau }^{\mu }v$ with $\tau (p)>0$ and $v(p)>2$. In a neighborhood with these strict inequalities put $f=\log v+\mu \log \tau$. By [F5], ${\tau }^{\mu }>0$ and $f=\log ({\tau }^{\mu }v)$ there. By [F4], $f$ is smooth and has a local maximum at $p$. Apply the Euclidean extremum test in [F3] to its coordinate representation in an interior chart: $df(p)=0$ and the matrix $({\mathrm{\partial }}_i{\mathrm{\partial }}_{j}f)(p)$ is negative semidefinite. The connection term in the coordinate Hessian formula vanishes because $df(p)=0$. Thus ${\mathrm{Hess}}_{h}f(p)$ is negative semidefinite, ${\mathrm{\nabla }}_{\mathrm{\Sigma }}f(p)=0$, and ${\mathrm{\Delta }}_{\mathrm{\Sigma }}f(p)={\sum }_{a=1}^3{\mathrm{Hess}}_{h}f(e_a,e_a)\le 0$ for an $h_p$-orthonormal basis $(e_a)$.

By the coordinate scalar rules and the Hessian and trace formulas in [F3] and [F4], ${\mathrm{\nabla }}_{\mathrm{\Sigma }}\log \tau ={\mathrm{\nabla }}_{\mathrm{\Sigma }}\tau \mathrm{/}\tau$ and ${\mathrm{\Delta }}_{\mathrm{\Sigma }}\log \tau ={\mathrm{\Delta }}_{\mathrm{\Sigma }}\tau \mathrm{/}\tau -\mathrm{\mid }{\mathrm{\nabla }}_{\mathrm{\Sigma }}\tau {\mathrm{\mid }}^2\mathrm{/}{\tau }^2$. At $p$, step 1.1 therefore gives ${\mathrm{\nabla }}_{\mathrm{\Sigma }}\log v=-\mu {\mathrm{\nabla }}_{\mathrm{\Sigma }}\tau \mathrm{/}\tau$. Substituting this into the assumed inequalities gives $0\ge {\mathrm{\Delta }}_{\mathrm{\Sigma }}f\ge ({\mu }^2\mathrm{/}4-\mu )\mathrm{\mid }{\mathrm{\nabla }}_{\mathrm{\Sigma }}\tau {\mathrm{\mid }}^2\mathrm{/}{\tau }^2-C{v}^2(1+\mu \mathrm{/}\tau )$.

Multiply by ${\tau }^2>0$ and use [F1]. Put $A=T^2+\mu T>0$. The lower bound from step 2.1 is at least $2CA(v^2-1)-C({\tau }^2+\mu \tau )v^2\ge CA(v^2-2)>0$, since $0<\tau \le T$ and $v>2$. This contradicts $\mathrm{\Delta }f\le 0$. Such a $\mu$ exists because the left side of its required inequality is quadratic in $\mu$ and the right side is affine.