network profile
| bio | website | |
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| location | ||
| age | ||
| visits | member for | 5 months |
| seen | Jan 11 at 21:34 | |
| stats | profile views | 16 |
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Dec 15 |
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Absolutely continuous functions with derivatives in $L^p$ @Jonas. Yes you are totally right. It is integrable on bounded intervals, and this is what I use for the proof. Sorry for my confusion! |
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Dec 15 |
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Absolutely continuous functions with derivatives in $L^p$ @Jonas. Yes thank you! |
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Dec 15 |
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Absolutely continuous functions with derivatives in $L^p$ ...is integrable, and we have $F(b)-F(a)= \int_a^b F'(x)dx$. I will post an answer to my question so that you will understand the discussion between Giuseppe and I. |
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Dec 15 |
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Absolutely continuous functions with derivatives in $L^p$ @Jonas: The problem comes from a problems set that our professor gave us to practise for the final exam. I am not sure where she took them from. The definition of absolutely continuous function that I use is: For every $\epsilon>0$, there exists $\delta > 0$ such that for any finite collection of disjoint intervals $(a_i, b_i)_{i=1}^n$, $\sum_i |F(b_i)-F(a_i)|<\epsilon$ whenever $\sum_i (b_i-a_i) < \delta$. This is the definition from Folland's real analysis. The Fundamental theorem of calculus states that $F$ is absolutely continuous if and only if $F'$ exists almost everywhere and... |
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Dec 14 |
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Absolutely continuous functions with derivatives in $L^p$ @Jonas. Is it important in this case? |
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Dec 14 |
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Absolutely continuous functions with derivatives in $L^p$ @Jonas. It is not directly part of my definition, but it follows from the fundamental theorem of calculus for Lebesgue integration. |
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Dec 12 |
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Absolutely continuous functions with derivatives in $L^p$ Oh if you take $L = \int_{\mathbb R} f'(t) dt$, then I think it works applying Holder. |
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Dec 12 |
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Absolutely continuous functions with derivatives in $L^p$ I thought about that. Then doesn't the constant $L$ that you get depend on $x$ and $y$? |
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Dec 11 |
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How to know a function is in $L^p$. Oh sorry about my confusion! Thank you! |
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Dec 11 |
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How to know a function is in $L^p$. I am not sure to see why we get what we want. We don't know anything about $m(\{x | f(x)^p > t\}$. It seems to me that you have only proven the well-known identity $\int g(x) = \int m(\{x | g(x) > t\}$ for any function $g$. |
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Dec 11 |
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On an identity about integrals I see. Thank you, I did not realize that! |
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Dec 10 |
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On an identity about integrals Thanks! The way you put it is really intuitive! |
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Dec 10 |
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On one property of the Lebesgue Measure @Nate. I am not sure I see how the answers address this question. What do you mean by taking complements? |
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Dec 10 |
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On one property of the Lebesgue Measure @Jacob. $I$ is an arbitrary measurable subset of $[0,1]$. |
