We keep, until the end of this third part, the hypotheses and notation of the previous question. For $x , y \in I$ such that $y \neq x$, we set $$H _ { f } ( x , y ) = \frac { x f ( y ) - y f ( x ) } { f ( y ) - f ( x ) }$$ (a) Show that for all $x , y \in I$ such that $y \neq x$ we have $$H _ { f } ( x , y ) = x - f ( x ) \int _ { 0 } ^ { 1 } g ^ { \prime } ( \lambda f ( x ) + ( 1 - \lambda ) f ( y ) ) d \lambda$$ (b) Deduce that $H _ { f }$ admits a unique continuous extension to $I \times I$ as a whole. We still denote this extension by $H _ { f } : I \times I \rightarrow \mathbb { R }$. (c) Show that $H _ { f }$ is of class $\mathcal { C } ^ { 2 }$ on $I \times I$. (d) Compute $H _ { f } ( x , x )$.
We keep, until the end of this third part, the hypotheses and notation of the previous question. For $x , y \in I$ such that $y \neq x$, we set
$$H _ { f } ( x , y ) = \frac { x f ( y ) - y f ( x ) } { f ( y ) - f ( x ) }$$
(a) Show that for all $x , y \in I$ such that $y \neq x$ we have
$$H _ { f } ( x , y ) = x - f ( x ) \int _ { 0 } ^ { 1 } g ^ { \prime } ( \lambda f ( x ) + ( 1 - \lambda ) f ( y ) ) d \lambda$$
(b) Deduce that $H _ { f }$ admits a unique continuous extension to $I \times I$ as a whole. We still denote this extension by $H _ { f } : I \times I \rightarrow \mathbb { R }$.
(c) Show that $H _ { f }$ is of class $\mathcal { C } ^ { 2 }$ on $I \times I$.
(d) Compute $H _ { f } ( x , x )$.