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Let $f\colon [0,1]\rightarrow \mathbb{R}$ be a bounded function. Let $P_1=[0,\frac{1}{2}]\cup [\frac{1}{2},1]$ and $P_2=[0,\frac{1}{3}]\cup [\frac{1}{3}, \frac{2}{3}]\cup [\frac{2}{3},1]$ be partitions of $[0,1]$. Corresponding to each partition, we define the lower sums:

$L_1(f)=\frac{1}{2}m_1+\frac{1}{2}m_2$, where $m_1,m_2$ are minimum values of $f$ in the intervals $[0,\frac{1}{2}], [\frac{1}{2}, 1]$ respectively. Similarly, we define the lower sum $L_2$ for $P_2$. I will fix these partitions for my question.

Here, partitions $P_1$ and $P_2$ are not refinements of each other, so I can not say anything about comparison between $L_(f)$ and $L_2(f)$.

Question: Does there exists a bounded function $f\colon [0,1]\rightarrow \mathbb{R}$ such that $L_1(f)>L_2(f)$?

I don't know answer to this problem, and I tried to construct such $f$, but I failed. Even, I don't know whether it is impossible that $L_1(f)>L_2(f)$ for any function $f$ with prescribed partitions $P_1$ and $P_2$ above. Please help me.

Sorry! I forgot to add one more requirement on $f$, we would consider $f$ to be non-negative, i.e. $f(x)\geq 0$ for all $x$.

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Take $f(x) = 0$ on $x \in [0,0.5]$ and $f(x)=1$ on $x \in [0.5,1]$.

$L_1(f)=\tfrac{1}{2}$ and $L_2(f)=\tfrac{1}{3}$

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  • $\begingroup$ Btw, my function $f$ is not continuous, but because you can have a series of $C^{\infty}$ functions which converge uniformly towards $f$ (e.g. polynoms - Weierstrass theorem), the result obviously holds for $C^{\infty}$ functions $\endgroup$ – Alexandre Halm Sep 29 '14 at 7:41
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Hint. Consider step functions which are constant on $[0,1/2]$ and $[1/2,1]$.

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  • $\begingroup$ Oh! Thanks! Simple example. Nice. $\endgroup$ – Groups Sep 29 '14 at 10:35

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