Proof Analysis: Can You Check My Work and Provide Feedback?

[Artusartos]

Can anybody tell me if my proof is correct? If not, can you give me a hint? I attached my answer...

Thanks in advance

 

[pasmith]

Your proof doesn't work: at one point you conclude that $U(f^2,P) - L(f^2,P) = 0$ which is not generally the case: consider f(x) = x on [a,b].

 

[Artusartos]

Your proof doesn't work: at one point you conclude that $U(f^2,P) - L(f^2,P) = 0$ which is not generally the case: consider f(x) = x on [a,b].

Thanks, but if $$U(f^2,P) \leq B^2(b-a)$$ and $$L(f^2,P) \geq B^2(b-a)$$ then why aren't they equal?

 

[pasmith]

Thanks, but if $$U(f^2,P) \leq B^2(b-a)$$ and $$L(f^2,P) \geq B^2(b-a)$$ then why aren't they equal?

Your second inequality isn't true. You should know that for any bounded function g, $L(g,P) \leq U(g,P) \leq (b-a)\sup_{[a,b]} g$ and f^2 is a bounded function if f is.

 

[pasmith]

Here's a hint: $\sup f^2 = (\sup |f|)^2$ and $\inf f^2 = (\inf |f|)^2$ (why?). Therefore, on each interval of the partition,
$$\sup f^2 - \inf f^2 = (\sup |f|)^2 - (\inf |f|)^2 \\<br /> = (\sup |f| + \inf |f|)(\sup |f| - \inf |f|) \\<br /> \leq 2B(\sup |f| - \inf |f|).$$
Now you just have to show that on each interval of the partition,
$$\sup |f| - \inf |f| \leq \sup f - \inf f.$$
You'll have to classify intervals into three types: those where $f \geq 0$, those where $f \leq 0$, and those where f changes sign.