A function $f \in \mathcal{C}^2(\mathbb{R}^2, \mathbb{R})$ satisfies (1) on $\mathbb{R}^2$ if and only if $$\forall ( x , y ) \in \mathbb{R}^2, \quad \frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } ( x , y ) \times \frac { \partial ^ { 2 } f } { \partial y ^ { 2 } } ( x , y ) - \left( \frac { \partial ^ { 2 } f } { \partial x \partial y } ( x , y ) \right) ^ { 2 } = 1$$ We denote by $\mathcal { P } _ { 2 }$ the set of polynomial functions of degree $\leqslant 2$ from $\mathbb { R } ^ { 2 }$ to $\mathbb { R }$. If $( x _ { 0 } , y _ { 0 } )$ is in $\mathbb { R } ^ { 2 }$, show that there exists an open set $U$ of $\mathbb { R } ^ { 2 }$ containing $( x _ { 0 } , y _ { 0 } )$ such that the set of functions in $\mathcal { C } ^ { 2 } ( U , \mathbb { R } )$ satisfying (1) on $U$ and not coinciding on $U$ with any element of $\mathcal { P } _ { 2 }$ is infinite.
A function $f \in \mathcal{C}^2(\mathbb{R}^2, \mathbb{R})$ satisfies (1) on $\mathbb{R}^2$ if and only if
$$\forall ( x , y ) \in \mathbb{R}^2, \quad \frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } ( x , y ) \times \frac { \partial ^ { 2 } f } { \partial y ^ { 2 } } ( x , y ) - \left( \frac { \partial ^ { 2 } f } { \partial x \partial y } ( x , y ) \right) ^ { 2 } = 1$$
We denote by $\mathcal { P } _ { 2 }$ the set of polynomial functions of degree $\leqslant 2$ from $\mathbb { R } ^ { 2 }$ to $\mathbb { R }$.
If $( x _ { 0 } , y _ { 0 } )$ is in $\mathbb { R } ^ { 2 }$, show that there exists an open set $U$ of $\mathbb { R } ^ { 2 }$ containing $( x _ { 0 } , y _ { 0 } )$ such that the set of functions in $\mathcal { C } ^ { 2 } ( U , \mathbb { R } )$ satisfying (1) on $U$ and not coinciding on $U$ with any element of $\mathcal { P } _ { 2 }$ is infinite.