Assume f(x,y) and g(x,y) are two smooth functions on $R^2$
Let $$S=\left \{ (x,y)|f(x,y)=0 \right \} $$, and $$p=(a,b)\in S$$
such that
$$\frac{\partial f}{\partial x} (p) = -4,\space\frac{\partial f}{\partial y} (p) = 2,\space\frac{\partial g}{\partial x} (p) = 12,\space \frac{\partial g}{\partial y} (p)=-6$$
And
$$\begin{bmatrix} \frac{\partial^2}{\partial x^2}f(p) & \frac{\partial^2}{\partial x\partial y}f(p) \\ \frac{\partial^2}{\partial y\partial x}f(p) & \frac{\partial^2}{\partial x^2}f(p)\end{bmatrix}=\begin{bmatrix} 1 &2 \\ 2 &4\end{bmatrix}$$
$$\begin{bmatrix} \frac{\partial^2}{\partial x^2}g(p) & \frac{\partial^2}{\partial x\partial y}g(p) \\ \frac{\partial^2}{\partial y\partial x}g(p) & \frac{\partial^2}{\partial x^2}g(p)\end{bmatrix}=\begin{bmatrix} 3&-1 \\ -1 &2\end{bmatrix}$$
How to prove $ \exists \space R> 0 $ such that $\forall \space q \in {S}\cap\left \{(x,y)| x^2+y^2< R^2 \right \} $ there is always g(p)< g(q)
The only thing I am certain is we can prove this by implicit function theorem and Lagrange multiplier, but other than that, I have absolutely no clue where I should start. Could someone give me some hint?
By the implicit function theorem, there is a function $h$ such that $h(a)=b$ and $f(x,h(x))=0$ for all $x$ near $a$. If you differentiate and use the chain rule, you get $$\partial_xf(x,h(x))+\partial_yf(x,h(x)))h’(x)=0$$ In particular, if you plug in $x=a$, you get that $h’(a)=2$. You need to differentiate once more to compute $h’’(a)$.
Now consider the function $p(x)=g(x,h(x))$, which is the function $g$ restricted to the set $S$ near the point $p$. By Taylor’s formula of order two, you can write $$p(x)=p(a)+p’(a)(x-a)+p’’(a)(x-a)^2/2+o((x-a)^2)$$ By the chain rule, $$p’(x)=\partial_xg(x,h(x))+\partial_yg(x,h(x)))h’(x)$$ If you plug in $x=a$, you get $p’(a)=0$, so $a$ is a critical point of $p$. Now you need to differentiate once more to compute $p’’(a)$ and prove that it is positive. Good luck with all the chain rules and computations!