Let $k \in \mathbb { N } ^ { * }$. We denote by $c _ { k }$ the positive real number such that: $c _ { k } \int _ { 0 } ^ { 2 \pi } \left( \frac { 1 + \cos t } { 2 } \right) ^ { k } d t = 1$, and for every real $t$ we set $R _ { k } ( t ) = c _ { k } \left( \frac { 1 + \cos t } { 2 } \right) ^ { k }$. Let $\epsilon \in ] 0 , \pi [$, and $k \in \mathbb { N }$. Prove that for every $h \in C _ { 2 \pi } ( \mathbb { R } ; \mathbb { C } )$ that is of class $C ^ { 1 }$ on $\mathbb { R }$ and every real $u$, we have: $$\int _ { 0 } ^ { 2 \pi } R _ { k } ( u - t ) h ( t ) d t = \int _ { 0 } ^ { 2 \pi } R _ { k } \left( t _ { 1 } \right) h \left( u - t _ { 1 } \right) d t _ { 1 }$$ and $$\left| \int _ { 0 } ^ { 2 \pi } R _ { k } ( u - t ) h ( t ) d t - h ( u ) \right| \leq 2 \left\| h ^ { \prime } \right\| \epsilon + 4 \pi \| h \| d _ { k } ( \epsilon )$$ (We recall that $\int _ { 0 } ^ { 2 \pi } R _ { k } \left( t _ { 1 } \right) d t _ { 1 } = 1$ and that $\| h \|$ is defined at the beginning of the problem statement. To establish the inequality, one may use that $h \left( u - t _ { 1 } \right) = h \left( u - t _ { 1 } + 2 \pi \right)$ when $t _ { 1 } \in [ 2 \pi - \epsilon , 2 \pi ]$).
Let $k \in \mathbb { N } ^ { * }$. We denote by $c _ { k }$ the positive real number such that: $c _ { k } \int _ { 0 } ^ { 2 \pi } \left( \frac { 1 + \cos t } { 2 } \right) ^ { k } d t = 1$, and for every real $t$ we set $R _ { k } ( t ) = c _ { k } \left( \frac { 1 + \cos t } { 2 } \right) ^ { k }$.
Let $\epsilon \in ] 0 , \pi [$, and $k \in \mathbb { N }$. Prove that for every $h \in C _ { 2 \pi } ( \mathbb { R } ; \mathbb { C } )$ that is of class $C ^ { 1 }$ on $\mathbb { R }$ and every real $u$, we have:
$$\int _ { 0 } ^ { 2 \pi } R _ { k } ( u - t ) h ( t ) d t = \int _ { 0 } ^ { 2 \pi } R _ { k } \left( t _ { 1 } \right) h \left( u - t _ { 1 } \right) d t _ { 1 }$$
and
$$\left| \int _ { 0 } ^ { 2 \pi } R _ { k } ( u - t ) h ( t ) d t - h ( u ) \right| \leq 2 \left\| h ^ { \prime } \right\| \epsilon + 4 \pi \| h \| d _ { k } ( \epsilon )$$
(We recall that $\int _ { 0 } ^ { 2 \pi } R _ { k } \left( t _ { 1 } \right) d t _ { 1 } = 1$ and that $\| h \|$ is defined at the beginning of the problem statement. To establish the inequality, one may use that $h \left( u - t _ { 1 } \right) = h \left( u - t _ { 1 } + 2 \pi \right)$ when $t _ { 1 } \in [ 2 \pi - \epsilon , 2 \pi ]$).