We assume that $\frac { \sqrt { \lambda _ { 1 } } } { \sqrt { \lambda _ { 2 } } }$ is not a rational number. We fix an arbitrary element $f \in C _ { 2 \pi , 2 \pi } ^ { 1 } \left( \mathbb { R } ^ { 2 } ; \mathbb { C } \right)$. For each $k \in \mathbb { N } ^ { * }$ we set: $$\forall ( u , v ) \in \mathbb { R } ^ { 2 } , f _ { k } ( u , v ) = \int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 2 \pi } R _ { k } \left( u - \theta _ { 1 } \right) R _ { k } \left( v - \theta _ { 2 } \right) f \left( \theta _ { 1 } , \theta _ { 2 } \right) d \theta _ { 1 } d \theta _ { 2 }$$
Let $\epsilon \in ] 0 , \pi [$ and $k \in \mathbb { N } ^ { * }$. By writing $f _ { k } ( u , v ) - f ( u , v )$ as a sum of two terms and applying Question 10, prove that for every $( u , v ) \in \mathbb { R } ^ { 2 }$: $$\left| f _ { k } ( u , v ) - f ( u , v ) \right| \leq 2 \epsilon \left( \left\| \frac { \partial f } { \partial \theta _ { 1 } } \right\| + \left\| \frac { \partial f } { \partial \theta _ { 2 } } \right\| \right) + 8 \pi \| f \| d _ { k } ( \epsilon )$$