Let $n \in \mathbb { N } ^ { * }$ and $x \in [ 0,1 ]$. We consider the sum $$S ( x ) = \sum _ { k = 0 } ^ { n } \left| x - \frac { k } { n } \right| \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } .$$ We denote

• $V$ the set of integers $k \in \{ 0 , \ldots , n \}$ such that $\left| x - \frac { k } { n } \right| \leqslant \frac { 1 } { \sqrt { n } }$,
• $W$ the set of integers $k \in \{ 0 , \ldots , n \}$ such that $\left| x - \frac { k } { n } \right| > \frac { 1 } { \sqrt { n } }$, and we set

$$S _ { V } ( x ) = \sum _ { k \in V } \left| x - \frac { k } { n } \right| \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } \quad \text { and } \quad S _ { W } ( x ) = \sum _ { k \in W } \left| x - \frac { k } { n } \right| \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } .$$ a) Show that $S _ { V } ( x ) \leqslant \frac { 1 } { \sqrt { n } }$. b) Show that $S _ { W } ( x ) \leqslant \frac { x ( 1 - x ) } { \sqrt { n } }$. c) Deduce that $S ( x ) \leqslant \frac { 5 } { 4 \sqrt { n } }$.

Let $n \in \mathbb { N } ^ { * }$ and $x \in [ 0,1 ]$. We consider the sum $$S ( x ) = \sum _ { k = 0 } ^ { n } \left| x - \frac { k } { n } \right| \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } .$$ a) Write the Cauchy-Schwarz inequality in the space $\mathbb { R } ^ { n + 1 }$ equipped with its canonical inner product. b) Using question I.A.4, deduce that $S ( x ) \leqslant \frac { 1 } { 2 \sqrt { n } }$.

We denote by $\mathcal { C }$ the vector space of continuous functions from $[ 0,1 ]$ to $\mathbb { R }$, equipped with the supremum norm $\| f \| _ { \infty } = \sup _ { x \in [ 0,1 ] } | f ( x ) |$. For $f \in \mathcal { C }$ and $n \in \mathbb { N } ^ { * }$, the $n$-th Bernstein polynomial of $f$ is defined by $$B _ { n } ( f ) ( x ) = \sum _ { k = 0 } ^ { n } f \left( \frac { k } { n } \right) \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k }$$ for all $x \in [ 0,1 ]$.
If $f ( x ) = x ^ { 2 }$ for all $x \in [ 0,1 ]$, determine, for all $n \in \mathbb { N } ^ { * }$, the polynomial $B _ { n } ( f )$ and deduce the value of $\left\| B _ { n } ( f ) - f \right\| _ { \infty }$.

We denote by $\mathcal { C }$ the vector space of continuous functions from $[ 0,1 ]$ to $\mathbb { R }$, equipped with the supremum norm $\| f \| _ { \infty } = \sup _ { x \in [ 0,1 ] } | f ( x ) |$. For $f \in \mathcal { C }$ and $n \in \mathbb { N } ^ { * }$, the $n$-th Bernstein polynomial of $f$ is defined by $$B _ { n } ( f ) ( x ) = \sum _ { k = 0 } ^ { n } f \left( \frac { k } { n } \right) \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k }$$ for all $x \in [ 0,1 ]$.
Let $f \in \mathcal { C }$. Show, for all $x \in [ 0,1 ]$, the relation $$B _ { n } ( f ) ( x ) - f ( x ) = \sum _ { k = 0 } ^ { n } \left( f \left( \frac { k } { n } \right) - f ( x ) \right) \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k }$$

We denote by $\mathcal { C }$ the vector space of continuous functions from $[ 0,1 ]$ to $\mathbb { R }$, equipped with the supremum norm $\| f \| _ { \infty } = \sup _ { x \in [ 0,1 ] } | f ( x ) |$. For $f \in \mathcal { C }$ and $n \in \mathbb { N } ^ { * }$, the $n$-th Bernstein polynomial of $f$ is defined by $$B _ { n } ( f ) ( x ) = \sum _ { k = 0 } ^ { n } f \left( \frac { k } { n } \right) \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k }$$ for all $x \in [ 0,1 ]$.
a) Show that if $f$ is $\delta$-Lipschitz, then $\left\| B _ { n } ( f ) - f \right\| _ { \infty } \leqslant \frac { \delta } { 2 \sqrt { n } }$ for all integer $n \geqslant 1$. b) Deduce that if $f$ is of class $C ^ { 1 }$, then there exists a real $c$ such that, for all $n \in \mathbb { N } ^ { * } , \left\| B _ { n } ( f ) - f \right\| _ { \infty } \leqslant \frac { c } { \sqrt { n } }$. c) Extend the previous result to the case where $f$ is a continuous function, piecewise $C ^ { 1 }$.

We denote by $\mathcal { C }$ the vector space of continuous functions from $[ 0,1 ]$ to $\mathbb { R }$, equipped with the supremum norm $\| f \| _ { \infty } = \sup _ { x \in [ 0,1 ] } | f ( x ) |$. For $f \in \mathcal { C }$ and $n \in \mathbb { N } ^ { * }$, the $n$-th Bernstein polynomial of $f$ is defined by $$B _ { n } ( f ) ( x ) = \sum _ { k = 0 } ^ { n } f \left( \frac { k } { n } \right) \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k }$$ for all $x \in [ 0,1 ]$.
Let $f : [ 0,1 ] \rightarrow \mathbb { R }$ be a continuous function, piecewise $C ^ { 1 }$. Deduce from the above that, for all real $r > 0$, there exists a polynomial $P$ with real coefficients such that $\forall x \in [ 0,1 ] , f ( x ) - r \leqslant P ( x ) \leqslant f ( x ) + r$.

Let $g : [ 0,1 ] \rightarrow \mathbb { R }$ be the function such that $g ( x ) = 1 / x$ if $x \geqslant \mathrm { e } ^ { - 1 }$ and $g ( x ) = 0$ otherwise. We fix a real $\varepsilon \in ] 0 , \mathrm { e } ^ { - 1 } [$. We define two continuous applications $g ^ { + } , g ^ { - } : [ 0,1 ] \rightarrow \mathbb { R }$ as follows:

• $g ^ { + }$ is affine on $\left[ \mathrm { e } ^ { - 1 } - \varepsilon , \mathrm { e } ^ { - 1 } \right]$ and coincides with $g$ on $\left[ 0 , \mathrm { e } ^ { - 1 } - \varepsilon \right] \cup \left[ \mathrm { e } ^ { - 1 } , 1 \right]$;
• $g ^ { - }$ is affine on $\left[ \mathrm { e } ^ { - 1 } , \mathrm { e } ^ { - 1 } + \varepsilon \right]$ and coincides with $g$ on $\left[ 0 , \mathrm { e } ^ { - 1 } \left[ \cup \left[ \mathrm { e } ^ { - 1 } + \varepsilon , 1 \right] \right. \right.$.

Establish the existence of two polynomials $P , Q$ with real coefficients such that: $$\forall x \in [ 0,1 ] , \quad g ^ { - } ( x ) - \varepsilon \leqslant P ( x ) \leqslant g ( x ) \leqslant Q ( x ) \leqslant g ^ { + } ( x ) + \varepsilon$$