A function $f \in \mathcal{C}^2(\Omega, \mathbb{R})$ satisfies (1) on $\Omega$ if and only if $$\forall ( x , y ) \in \Omega, \quad \frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } ( x , y ) \times \frac { \partial ^ { 2 } f } { \partial y ^ { 2 } } ( x , y ) - \left( \frac { \partial ^ { 2 } f } { \partial x \partial y } ( x , y ) \right) ^ { 2 } = 1$$ Let $\Omega$ be a non-empty open set of $\mathbb { R } ^ { 2 }$, $f$ in $\mathcal { C } ^ { 2 } ( \Omega , \mathbb { R } )$ satisfying (1) on $\Omega$, $( a , b ) \in \mathbb { R } ^ { 2 }$, $\Omega _ { a , b }$ the image of $\Omega$ by the translation of vector $( a , b )$ and $f _ { a , b }$ the function defined on $\Omega _ { a , b }$ by $$\forall ( x , y ) \in \Omega _ { a , b } , \quad f _ { a , b } ( x , y ) = f ( x - a , y - b )$$ Show that $f _ { a , b }$ satisfies (1) on $\Omega _ { a , b }$.
A function $f \in \mathcal{C}^2(\Omega, \mathbb{R})$ satisfies (1) on $\Omega$ if and only if
$$\forall ( x , y ) \in \Omega, \quad \frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } ( x , y ) \times \frac { \partial ^ { 2 } f } { \partial y ^ { 2 } } ( x , y ) - \left( \frac { \partial ^ { 2 } f } { \partial x \partial y } ( x , y ) \right) ^ { 2 } = 1$$
Let $\Omega$ be a non-empty open set of $\mathbb { R } ^ { 2 }$, $f$ in $\mathcal { C } ^ { 2 } ( \Omega , \mathbb { R } )$ satisfying (1) on $\Omega$, $( a , b ) \in \mathbb { R } ^ { 2 }$, $\Omega _ { a , b }$ the image of $\Omega$ by the translation of vector $( a , b )$ and $f _ { a , b }$ the function defined on $\Omega _ { a , b }$ by
$$\forall ( x , y ) \in \Omega _ { a , b } , \quad f _ { a , b } ( x , y ) = f ( x - a , y - b )$$
Show that $f _ { a , b }$ satisfies (1) on $\Omega _ { a , b }$.