Hi in the proof of f integrable, $g(x)= \int_{[0,x]} f dt$ is uniformly continuous, i saw a proof that uses the following argument. Let $\epsilon > 0$ be given, take $δ := \epsilon^-1 · (\int |f|dt) >0$ since f is integrable. Let $x My question is, how do we get that $\int_{[x,y]} |f(t)|dt ≤ |x − y|\int_{R}
|f(t)|dt$? I have tried to bound it and only able to conclude that $\int_{[x,y]} |f(t)|dt\leq |x − y| sup(|f|)$ ? Thanks in advance
Why is $\int_{[x,y]}| f |\leq |x-y| \int_{R}| f| $?
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integration
analysis
measure-theory
lebesgue-integral
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2The proposed inequality is false. That would imply that $g$ is Lipschitz on $\Bbb{R}$, which then implies that $f$ is bounded. Of course, there are integrable functions which are unbounded on any open sets. This necessitates the control over such exceptional 'peaks'. – 2017-02-08
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0I see, so if f is bounded and integrable, then we can do this? – 2017-02-08
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0@Socchi: No, even then it is false. Take $f = 2 \cdot 1_{[0,1/2]}$. – 2017-02-08
1 Answers
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You can't, the proposed proof is wrong. A simple counterexample is $f(x) = |x|^{-1/2}$ for $-1 \le x \le 1$, $f(x)=0$ otherwise. Note $g(x) = 2 \sqrt{|x|}$ for $-1 \le x \le 1$ which is not Lipschitz.
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0But your f(x) is not integrable? – 2017-02-08
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0I was worried about the region, thanks. I wonder whether a part of the proof is missing? – 2017-02-08
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0Oh, at infinity, you're right. Fixed. What's relevant is what happens near 0. – 2017-02-08
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0Thanks, I guess that claim is wrong? Maybe some part of the proof is missing to get to that conclusion? – 2017-02-08
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0I think it's just wrong. I can't think of any useful class of functions $f$ for which it is true. – 2017-02-08