We assume that $f \in \mathcal { C } ( [ 0,1 ] , \mathbb { R } )$ satisfies: $$\exists \alpha \in ]0,1] , \exists K \geq 0 , \forall ( y , z ) \in [ 0,1 ] ^ { 2 } , | f ( y ) - f ( z ) | \leq K | y - z | ^ { \alpha }$$ For all $n \in \mathbb{N}^*$, define $B _ { n } f ( X ) = \sum _ { k = 0 } ^ { n } f \left( \frac { k } { n } \right) \binom { n } { k } X ^ { k } ( 1 - X ) ^ { n - k }$. Let $n \in \mathbb { N } ^ { * }$. Show that $$\left\| f - B _ { n } f \right\| _ { \infty } \leq \frac { 3 K } { 2 } \frac { 1 } { n ^ { \alpha / 2 } }$$ Hint: One may first express $f ( x ) - B _ { n } f ( x )$ in terms of $\mathbb { E } ( f ( x ) - f \left( S _ { n } \right) )$.
We assume that $f \in \mathcal { C } ( [ 0,1 ] , \mathbb { R } )$ satisfies:
$$\exists \alpha \in ]0,1] , \exists K \geq 0 , \forall ( y , z ) \in [ 0,1 ] ^ { 2 } , | f ( y ) - f ( z ) | \leq K | y - z | ^ { \alpha }$$
For all $n \in \mathbb{N}^*$, define $B _ { n } f ( X ) = \sum _ { k = 0 } ^ { n } f \left( \frac { k } { n } \right) \binom { n } { k } X ^ { k } ( 1 - X ) ^ { n - k }$.
Let $n \in \mathbb { N } ^ { * }$. Show that
$$\left\| f - B _ { n } f \right\| _ { \infty } \leq \frac { 3 K } { 2 } \frac { 1 } { n ^ { \alpha / 2 } }$$
Hint: One may first express $f ( x ) - B _ { n } f ( x )$ in terms of $\mathbb { E } ( f ( x ) - f \left( S _ { n } \right) )$.