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Suppose $f:[a,b]\to \mathbb{R}$ is Lipschitz, i.e $|f(x)-f(y)| \leq C|x-y|$ for all $x,y$ in $[a,b]$ and thus $f$ is continuous. Show that for any partition $P$ of $[a,b]$, $U(f,P) - L(f,P) \leq C(b-a)mesh(P)$.

Some useful facts:

Partition of $[a,b]$ is $P=\{a=x_0 < x_1 < ... < x_{n-1} < x_n\}$
$\Delta_j = x_j - x_{j-1}$
$mesh(P)= \max\{\Delta_j : 1 \leq j\leq n\}$.

$U(f,P) = \sum_{j=1}^n \sup\{f(x): x_{j-1} \leq x \leq x _j\} \cdot \Delta _j$
$L(f,P) = \sum_{j=1}^n \inf\{f(x): x_{j-1} \leq x \leq x_j\} \cdot \Delta_j$

Lemma: If $P$ and $Q$ are partitions of $[a,b]$, then $L(f,P) \leq U(f,Q)$.

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    $\begingroup$ So you won't ever use LaTeX to write your questions...? $\endgroup$ – DonAntonio Apr 3 '13 at 14:47
  • $\begingroup$ sorry, Ill try to fix it somemore $\endgroup$ – sarah Apr 3 '13 at 14:56
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Let $I=[c,d] \subset [a,b]$. Then $f(x)-f(y) \le |f(x)-f(y)| \le C |x-y| \le C (d-c)$, for all $x,y \in I$. It follows that $\sup_{x\in I} f(x) - \inf_{x\in I} f(y) \le C (d-c)$.

Hence if $I_j=[x_{j-1},x_j]$, we have $\sup_{x\in I_j} f(x) - \inf_{x\in I_j} f(y) \le C \Delta_j$.

Then you have $U(f,P)-L(f,P) \le \sum_j C \Delta_j \Delta_j$. Replace one of the $\Delta_j$ by $\operatorname{mesh} P$ to get $U(f,P)-L(f,P) \le (\sum_j C \Delta_j ) \operatorname{mesh} P$, and note that $\sum_j \Delta_j = b-a$ to get $U(f,P)-L(f,P) \le C(b-a) \operatorname{mesh} P$.

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  • $\begingroup$ why is |x-y| $\leq$ (d-c)? thanks $\endgroup$ – sarah Apr 3 '13 at 15:09
  • $\begingroup$ Because $x,y \in [c,d]$. $\endgroup$ – copper.hat Apr 3 '13 at 15:12
  • $\begingroup$ thanks, do we need the interval [c,d]? couldnt we just use [a,b]? $\endgroup$ – sarah Apr 3 '13 at 15:17
  • $\begingroup$ No, you need it for the interval $[x_{j-1},x_j]$ (since this is the interval over which the $\sup$ and $\inf$ are taken). $\endgroup$ – copper.hat Apr 3 '13 at 16:44

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