We denote by $C _ { 2 \pi , 2 \pi } \left( \mathbb { R } ^ { 2 } ; \mathbb { C } \right)$ the set of continuous functions $f : \mathbb { R } ^ { 2 } \rightarrow \mathbb { C }$ such that: $$\forall \left( \theta _ { 1 } , \theta _ { 2 } \right) \in \mathbb { R } ^ { 2 } , f \left( \theta _ { 1 } + 2 \pi , \theta _ { 2 } \right) = f \left( \theta _ { 1 } , \theta _ { 2 } \right) = f \left( \theta _ { 1 } , \theta _ { 2 } + 2 \pi \right)$$
Let $f \in C _ { 2 \pi , 2 \pi } \left( \mathbb { R } ^ { 2 } ; \mathbb { C } \right)$. Prove that $$\sup _ { \left( \theta _ { 1 } , \theta _ { 2 } \right) \in \mathbb { R } ^ { 2 } } \left| f \left( \theta _ { 1 } , \theta _ { 2 } \right) \right| = \sup _ { \left( \theta _ { 1 } , \theta _ { 2 } \right) \in [ 0,2 \pi ] ^ { 2 } } \left| f \left( \theta _ { 1 } , \theta _ { 2 } \right) \right|$$ Deduce that $\left( \theta _ { 1 } , \theta _ { 2 } \right) \mapsto \left| f \left( \theta _ { 1 } , \theta _ { 2 } \right) \right|$ is bounded on $\mathbb { R } ^ { 2 }$ and attains its supremum.