We assume that $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ is a decreasing sequence of strictly positive real numbers. We denote by $f$ the step function which, for all $k \in \mathbb { N } ^ { * }$, equals $a _ { k }$ on the interval $[ k - 1 , k [$. Let $k$ be in $\mathbb { N } ^ { * }$. Prove that the function $v _ { k }$ defined on $[ k - 1 , k ]$ by $$\left\{ \begin{array} { l }
v _ { k } ( x ) = \frac { 1 } { x } \sum _ { i = 1 } ^ { k - 1 } \ln \left( a _ { i } \right) + \frac { 1 } { x } ( x - k + 1 ) \ln \left( a _ { k } \right) \quad \text { if } k \geqslant 2 \\
v _ { 1 } ( x ) = \ln \left( a _ { 1 } \right)
\end{array} \right.$$ is minimal for $x = k$.
We assume that $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ is a decreasing sequence of strictly positive real numbers. We denote by $f$ the step function which, for all $k \in \mathbb { N } ^ { * }$, equals $a _ { k }$ on the interval $[ k - 1 , k [$.
Let $k$ be in $\mathbb { N } ^ { * }$. Prove that the function $v _ { k }$ defined on $[ k - 1 , k ]$ by
$$\left\{ \begin{array} { l }
v _ { k } ( x ) = \frac { 1 } { x } \sum _ { i = 1 } ^ { k - 1 } \ln \left( a _ { i } \right) + \frac { 1 } { x } ( x - k + 1 ) \ln \left( a _ { k } \right) \quad \text { if } k \geqslant 2 \\
v _ { 1 } ( x ) = \ln \left( a _ { 1 } \right)
\end{array} \right.$$
is minimal for $x = k$.