I would like to know, under which simple conditions on $f : [0,1] \longmapsto \mathbb{R}$ a continuous function, the Riemann regular lower sum, defined to be : $$ L_n(f) := \frac{1}{n}\sum_{k = 0}^{n-1} f\left(\frac{k}{n}\right) $$ is non-decreasing as $n\longrightarrow +\infty$.
A well known result states that if $f$ is nondecreasing and $P'$ is a refinement of $P$, then $L(f,P') \geq L(f,P)$, where $L(f,Q)$ is the lower sum over the subdivision $Q$. This allows to say that $L_{2n}(f) - L_n(f) \geq 0$.
However do we have $L_{n+1}(f) - L_n(f)\geq 0$ whenever $f$ is nondecreasing ?
Every numerical explorations I have done so far outputs an increasing sequence $(L_n(f))$. I have tested both concave and convex functions hoping to discriminate these two properties.
When doing the algebraic calculations, nothing simple comes out.
A sketch of a proof or a counterexample would be greatly appreciated ! Thanks.
Side note : There seem to be an interesting convergence of the sequence $n(n+1)(L_{n+1}(f) - L_n(f))$.
Consider $f(x)=1$ for $x\geq 1/2,$ $10(x-2/5)$ for $x\in [2/5, 1/2)$ and 0 otherwise. This is continuous. Then $L_2=1/2$ and $L_3=1/3$.