Monotonicity of function averages

Question

Please let me know if you know an answer to this problem. May be you could provide a reference to some publication on this topic?

Let $f(x)$ be a real-valued strictly convex function on $[0, 1]$. For any integer $k$ between $0$ and some positive integer $n$ let $x_k =k/n$. Consider the average value function $g(n)= \sum_{k=0}^{n}f(x_k)/(n+1) $. Is it true that $g(n)$ is decreasing?

Thank you!

Answer 1

$g(n-1) \geq g(n) \Leftrightarrow (n+1) \sum f( \frac{i}{n-1}) \leq n \sum f( \frac{i}{n})$.

This follows by summing up the following inequlities:
$\begin{array} { l l l l l l l } & & n \times f(0) & = & n \times f(0) &=& n \times f(0), \\ 1 \times f(\frac{0}{n-1}) &+& (n-1) \times f( \frac{1}{n-1} ) & \geq & n \times f( \frac{\frac{0}{n-1} + \frac{n-1}{n-1}}{n}) &=& n f(\frac{1}{n}), \\ 2 \times f( \frac{1}{n-1}) &+& (n-2) \times f( \frac{2}{n-1}) &\geq& n \times f ( \frac{\frac{ 2}{n-1} + \frac{2n-4}{n-1}}{n}) &=& n f(\frac{2}{n}), \\ \vdots \\ i \times f( \frac{i-1}{n-1}) & + & (n-i) \times f( \frac{i}{n-1}) & \geq & n \times f( \frac{\frac{ i^2 - i}{n-1} + \frac{ni - i^2}{n-1} } { n } ) & = & n f(\frac{i}{n}), \\ (n-1) \times f( \frac{n-2}{n-1} ) &+& 1 \times f( \frac{n-1}{n-1}) &\geq& n \times f(\frac{ \frac{ n^2-3n+2}{n-1} + \frac{n-1}{n-1} }{n}) &=& n f(\frac{n-1}{n}) \\ n \times f(1) & & &=& n \times f(1)&=& n \times f(1) \\ \end{array}$

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Note: Strict convexity gives strict inequalities, so the function is strictly decreasing.

Answer 2

The following related monotonicity facts are also true: If, in addition, $f(x)$ is decreasing, then the sequence $u(n)= \sum_{k=0}^{n-1}f(x_k)/n$ is decreasing and the sequence $v(n)= \sum_{k=1}^{n}f(x_k)/n$ is increasing.

For the affirmative answer to the original problem this additional assumption about $f(x)$ is not needed.