If $f:x \to 2x+1$ over the interval $[1,3]$ and $\mathcal P$ be a partition consisting of the points $\{1,\frac32,2,3\} $, how do I find a partition $\mathcal P'$ of $[1,3]$ for which $U(f,\mathcal P')-L(f,\mathcal P') \lt 2 ?$
$U$ and $L$ denote the Upper and Lower sums of $f$

I don't even know where to start. So could you please help me out.

Any help is Much Appreciated!

Thanks in advance!

  • $\begingroup$ Trying some partitions and calculating their upper and lower sums seems like a good place to start. $\endgroup$ – Chris Eagle Mar 27 '12 at 9:31
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    $\begingroup$ How does $\mathcal P$ come into play? Do you need your $\mathcal P'$ be a refinement of $\mathcal P$? $\endgroup$ – martini Mar 27 '12 at 9:59
  • $\begingroup$ @martini I am not totally sure if $\mathcal P'$ is actually a refinement of $\mathcal P$ or just another partition in $[1,3]$. It's not very clear to me either but $\mathcal P'$ may not contain all the points in P.(or might it?) $\endgroup$ – funktor Mar 27 '12 at 10:22
  • $\begingroup$ But suppose that $\mathcal P'$ is a refinement of $\mathcal P$, what do i do in that case? $\endgroup$ – funktor Mar 27 '12 at 10:24
  • $\begingroup$ Did you try, what Chris proposes? I. e. compute $U(f,\mathcal P') - L(f, \mathcal P')$ for some $\mathcal P'$? If you need $\mathcal P \subseteq \mathcal P'$ you should of course try only such $\mathcal P'$. $\endgroup$ – martini Mar 27 '12 at 10:34

Observe that $U(f,\mathcal P)-L(f,\mathcal P)=3$. So, the partition you are looking for is preferably finer than the given partition $\cal P$. So, adding more points to $\cal P$ to form the partition $\cal P'$ is a way to go!

Why should this work? Let $P$ and $Q$ be two partitions of $[a,b]$. Suppose $P$ is finer than $Q$. (By slight abuse of notation, most authors also write $P \supseteq Q$.) Then, $$L(f,Q) \le L(f,P) \\U(f,Q) \ge U(f,P)$$

So, the inequalities would imply, $$U(f,\mathcal P')-L(f,\mathcal P') \le U(f,\mathcal P)-L(f, \mathcal P)$$

The slack inequality we show above does not prove that there actually exists a partition which achieves the upper bound that the question claims.

So, note that $f$ is monotonic and hence $f \in \scr R [a,b]$. So, $$\lim_{\|P\| \to 0}U(f,P)-L(f,P)=0$$

Given an $\varepsilon \gt 0,$ there exists at least one partition $P_ \varepsilon$ such that $$U(P_\varepsilon,f)-L(P_\varepsilon, f) \lt \varepsilon$$ Now, we are done showing the existence of a partition. Re-interpreting that limit, we must have that refining the partition, must eventually lead to a decrease of $U(f,P)-L(f,P)$ by any given positive quantity.

So, we must have a partition as sought by the question.

I'll only remark that $P_2=\left\{1,\dfrac 3 2, \dfrac 8 5, 2, \dfrac 5 2, 3\right\}$ works good enough, while I'll leave the details for you.

The general philosophy is, refining partitions increases the lower sum while it decreases the upper sum. It is what one would expect, if the function is integrable. From high school, we know that making finer "strips" will eventually give you the area under the curve.

Note: I am not claiming that it is the way to go, but one of the ways clearly seen given $\cal P$.

  • $\begingroup$ Ahh... made perfect sense. Thanks a lot $\endgroup$ – funktor Mar 27 '12 at 11:48
  • $\begingroup$ Glad to have been of some help. : ) $\endgroup$ – user21436 Mar 27 '12 at 11:49
  • $\begingroup$ @DidierPiau I had it right then. It is $3$. And, is the second question not atleast partially answered by "Why should it work?" section. I'd be glad to know what exactly needs to be added. $\endgroup$ – user21436 Mar 27 '12 at 18:56
  • $\begingroup$ So, I fixed the $3$ now. Thank you for pointing it out. If I am told, what needs to be there, I am most willing to add. $\endgroup$ – user21436 Mar 27 '12 at 18:57
  • $\begingroup$ Are you intending that I note: $f$ is monotonic and bounded and it should be integrable. So, refining partitions should eventually make the limit $$\lim_{\|P\| \to 0}[U(f,P)-L(f,P)]=0$$ $\endgroup$ – user21436 Mar 27 '12 at 18:59

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