Given level set show gradient is zero if function continuously differentiable.
I was asked to show if $\{(x,y)||x|=|y|\}$ is level set of continuously differentiable function $F(x,y)$ by implicit function theorem $\nabla F(0,0)=0$.
So if set is level set then $F(x,\pm x)=c$ for some constant $c$. Hint says assume $\nabla F(0,0)$ is not equal to zero and contradiction.
I know the theorem says if $O$ is open subset of $\mathbb{R}^2$ and $F$ is continuous differentiable, if $(x_0,y_0)$ is point in $O$ with $F(x_0,y_0)=0$ and $Df_y(x_0,y_0) \neq 0$ there is positive number $r$ and continuously differentiable function $g: I \rightarrow \mathbb{R}$ with $f(x,g(x))=0$ for all x in $I$, but don't know if that is correct implicit function theorem it called Dini theorem. How to start this?
I don't know if gradient not zero, one of partials is not zero. So if $\frac{\partial F}{\partial x}(0,0) \neq 0$. Then how to achieve contradiction? Or would be $F(0,0)=c \neq 0$?
Should I define function $g(x,y)=|x|-|y|$ then if gradient not zero, then both $Dg_x(0,0)$ and $Dg_y(0,0)$ not equal $0$?
Best Answer
1. The condition $F(x,\pm x)=c$ gives the partial derivatives in the diagonal directions: they are both zero, so the gradient is zero as well.
More formally, consider $G(u,v)=F(u+v,u-v)$, show that $D_uG(0,0)=D_vG(0,0)=0$, notice that $\nabla G = A \nabla F$ for some matrix $A$, and conclude.
2. The proof by contradiction, with the implicit function theorem, is less straightforward. Let us assume $\nabla F(0,0) \neq 0$. Then either $D_xF(0,0) \neq 0$ or $D_yF(0,0)\neq0$. Without loss of generality, we can assume $D_x F(0,0) \neq 0$. But then, the level set, in a neighbourhood of $(0,0)$, is the graph of a function $x \mapsto g(x)$. But it is not the graph of any function. Contradiction.