# Darboux sums inequality

Let $f$ be a continuous function on $[a,b]$, and continuously differentiable on $(a,b)$. Assume that $f'$ is bounded on $(a,b)$ and $\sup_{(a,b)}|f'(x)| = K$.

We'll also denote the upper and lower Darboux sums with respect to a partition $P$ of $[a,b]$ by $U(f, P)$ and $L(f, P)$ respectively. We need to show that for every partition of $[a,b]$ it holds that: $$0 \le U(f, P) - L(f, P) \le K(b-a)\Delta(P)$$

Where $\Delta(P)$ represents the diameter of $P$.

I tried using Lagrange's mean value thm, or uniform continuity but with no success.

I would appreciate your help with this question.

Thanks.

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How do you defined the diameter of a partition? Do you mean the mesh? Also, do you define the partition P to be $P=\{x_1,\cdots,x_n \}$ where $x_1<\cdots<x_n$ and $x_1=a$, $x_n=b$? – Pedro Tamaroff Apr 25 '12 at 23:19
$diam(P)=\Delta(P)=max_{1\le i \le n} x_{i}-x_{i-1}$. Partition of $[a,b]$ is a finite set of points that must contain $a$ and $b$ (like you wrote). – Amihai Zivan Apr 26 '12 at 6:54

Fix a partition $P=\{x_0,x_1,\ldots, x_n\}$ of $[a,b]$ with $\sup_i\{\Delta x_i\}\le \Delta P$, where $\Delta x_i=x_i-x_{i-1}$ for $i=1,2,\ldots,n$.
Then for each admissible $i$, consider the interval $I_i=[x_{i-1},x_i]$ determined by $P$: $f$ attains its maximum value over $I_i$ at some $x_i^M\in I_i$ and its minimum value over $I_i$ at some $x_i^m\in I_i$ with $x_i^m\ne x_i^M$. Then, by the Mean Value Theorem, for some $c_i$ between $x_i^M$ and $x_i^m$, we have $$f(x_i^M)-f(x_i^m)=|f(x_i^M)-f(x_i^m)| =|f'(c_i)||x_i^M-x_i^m|.$$
Now apply this to $$0\le U(f,P)-L(f,P)=\sum_{i=1}^n \bigl(\,f(x_i^M)-f(x_i^m)\,\bigr)\,\Delta x_i$$