We assume $c(x) = 0$ for all $x \in [0,1]$, $f \in \mathcal{C}([0,1],\mathbb{R})$ satisfies $|f(y)-f(z)| \leq K|y-z|^\alpha$ for some $\alpha \in ]0,1]$ and $K \geq 0$. Let $u$ be the solution of problem (1) and define $\hat{B}_{n+1}u$ as above. Let $n \in \mathbb { N } ^ { * }$ such that $n \geq 2$. We set $\chi _ { n + 1 } = \hat { B } _ { n + 1 } u - u$. (a) Show that $$\left\| \chi _ { n + 1 } ^ { \prime \prime } \right\| _ { \infty } \leq \left\| f - B _ { n - 1 } f \right\| _ { \infty } + \frac { 1 } { n + 1 } \| f \| _ { \infty } + K \frac { 1 } { ( n + 1 ) ^ { \alpha } }$$ (b) Show that for all $x \in [ 0,1 ]$ there exists $\xi \in [ 0,1 ]$ such that $$\chi _ { n + 1 } ( x ) = - \frac { 1 } { 2 } x ( 1 - x ) \chi _ { n + 1 } ^ { \prime \prime } ( \xi )$$ Hint: for $x \in ]0,1[$ one may consider the function $$h ( t ) = \chi _ { n + 1 } ( t ) - \frac { \chi _ { n + 1 } ( x ) } { x ( 1 - x ) } t ( 1 - t ) , \quad t \in [ 0,1 ]$$
We assume $c(x) = 0$ for all $x \in [0,1]$, $f \in \mathcal{C}([0,1],\mathbb{R})$ satisfies $|f(y)-f(z)| \leq K|y-z|^\alpha$ for some $\alpha \in ]0,1]$ and $K \geq 0$. Let $u$ be the solution of problem (1) and define $\hat{B}_{n+1}u$ as above. Let $n \in \mathbb { N } ^ { * }$ such that $n \geq 2$. We set $\chi _ { n + 1 } = \hat { B } _ { n + 1 } u - u$.
(a) Show that
$$\left\| \chi _ { n + 1 } ^ { \prime \prime } \right\| _ { \infty } \leq \left\| f - B _ { n - 1 } f \right\| _ { \infty } + \frac { 1 } { n + 1 } \| f \| _ { \infty } + K \frac { 1 } { ( n + 1 ) ^ { \alpha } }$$
(b) Show that for all $x \in [ 0,1 ]$ there exists $\xi \in [ 0,1 ]$ such that
$$\chi _ { n + 1 } ( x ) = - \frac { 1 } { 2 } x ( 1 - x ) \chi _ { n + 1 } ^ { \prime \prime } ( \xi )$$
Hint: for $x \in ]0,1[$ one may consider the function
$$h ( t ) = \chi _ { n + 1 } ( t ) - \frac { \chi _ { n + 1 } ( x ) } { x ( 1 - x ) } t ( 1 - t ) , \quad t \in [ 0,1 ]$$