We are given $f : \mathbb { R } ^ { 2 } \rightarrow \mathbb { C }$ and $g : \mathbb { R } ^ { 2 } \rightarrow \mathbb { R } ^ { 2 }$ two functions continuous and 1-periodic in each of their arguments, and $\omega \in \mathbb { R } ^ { 2 }$. We consider the system $$\left\{ \begin{array} { l }
F ^ { \prime } ( t ) = f ( \alpha ( t ) ) \\
\alpha ^ { \prime } ( t ) = \omega
\end{array} \right.$$ with $F(0)=0$, $\alpha(0)=(0,0)$. We assume that $f$ has zero average. The vector $\omega = \left( \omega _ { 1 } , \omega _ { 2 } \right)$ is said to be resonant if there exists $\left( k _ { 1 } , k _ { 2 } \right) \in \mathbb { Z } ^ { 2 } \backslash \{ ( 0,0 ) \}$ such that $k _ { 1 } \omega _ { 1 } + k _ { 2 } \omega _ { 2 } = 0$. Show that, if $\omega$ is resonant, there exists a function $f$ for which $F ( t ) = t$.
We are given $f : \mathbb { R } ^ { 2 } \rightarrow \mathbb { C }$ and $g : \mathbb { R } ^ { 2 } \rightarrow \mathbb { R } ^ { 2 }$ two functions continuous and 1-periodic in each of their arguments, and $\omega \in \mathbb { R } ^ { 2 }$. We consider the system
$$\left\{ \begin{array} { l }
F ^ { \prime } ( t ) = f ( \alpha ( t ) ) \\
\alpha ^ { \prime } ( t ) = \omega
\end{array} \right.$$
with $F(0)=0$, $\alpha(0)=(0,0)$. We assume that $f$ has zero average. The vector $\omega = \left( \omega _ { 1 } , \omega _ { 2 } \right)$ is said to be resonant if there exists $\left( k _ { 1 } , k _ { 2 } \right) \in \mathbb { Z } ^ { 2 } \backslash \{ ( 0,0 ) \}$ such that $k _ { 1 } \omega _ { 1 } + k _ { 2 } \omega _ { 2 } = 0$.
Show that, if $\omega$ is resonant, there exists a function $f$ for which $F ( t ) = t$.