Let $g$ be a function such that $g$ is $C^1$ on $\Bbb{R^3}$. Put $B=\{(x,y,z)\in \Bbb{R^3}\mid x^2+y^2+z^2\leq1\}$. since $g$ is continuous on $B$, we have $\max_{t\in B} g(t)=g(u)<\infty$.
Now if $u\in \{(x,y,z)\in \Bbb{R^3}\mid x^2+y^2+z^2<1\}$ then $\nabla g(u)=0$ (where $\nabla g(u)=(\frac{\partial g}{\partial x}(u),\frac{\partial g}{\partial y}(u),\frac{\partial g}{\partial z} (u))$. So there exists $a=0\in\Bbb R^3$ such that $\nabla g(u)=0 =au$. If $u\in\{(x,y,z)\in\Bbb R^3\mid x^2+y^2+z^2=1\}$ then by http://math.univ-lyon1.fr/~gelineau/devagreg/Extremas_lies.pdf $\exists a\in\Bbb R$ such that $\nabla g(u) =au$.
My question why $$a\geq 0$$.
Thanks .
We assume at $u$ the maximal value is assumed. Now if we have $\nabla g(u)=au$ with $a<0$, then we can show $g((1-\epsilon)u)>g(u)$ as follows: locally $g(u+\delta)=g(u)+\delta\cdot\nabla g(u)+O(\delta^2)$, so $g((1-\epsilon)u)=g(u)-\epsilon u \cdot \nabla g(u)+O(\epsilon^2)=g(u)-a(|u|^2)\epsilon +O(\epsilon^2)>g(u)$ when $\epsilon$ is sufficiently small. So prove by contradiction, we have $a\ge 0$.