# Can anybody check my proof (analysis)?

1. Dec 5, 2012

### Artusartos

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

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2. Dec 5, 2012

### 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].

3. Dec 5, 2012

### Artusartos

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?

4. Dec 5, 2012

### pasmith

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.

5. Dec 5, 2012

### 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,
$$\sup f^2 - \inf f^2 = (\sup |f|)^2 - (\inf |f|)^2 \\ = (\sup |f| + \inf |f|)(\sup |f| - \inf |f|) \\ \leq 2B(\sup |f| - \inf |f|).$$
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.