Consider the curve $C = \lbrace y=f(x)\rbrace$ in ${\bf R}^2$. Assume that $f$ is twice continuously differentiable. Then show that $m(C + C) \gt 0$ if and only if $C+C$ contains an open set, if and only if $f$ is not linear (where, presumably, $m$ is Lebesgue measure).
Real analysis by E. M. Stein: Chapter 1,Problem 7
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real-analysis
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0Sorry.This is the first time I use the website to ask questions,so something was wrong during writing them down.And my English is quite bad.I'm now learning how to use website properly. – 2011-08-12
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0Do you have thoughts you can share on the problem? Which implications do you already understand, and where are you stuck? – 2011-08-12
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1The page with that problem is not shown in the Google books preview. Nevertheless, I am going to try to edit some sense into the question. Wish me luck. – 2011-08-12
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1It may be worth noting that this problem is marked with an asterisk. "The ones that are most difficult, or go beyond the scope of the text, are marked with an asterisk." – 2011-08-12
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2"Consider the curve $\Gamma=\{y =f(x)\}$ in $\mathbb R^2$, $0\leq x \leq 1$.Assume that $f$ is twice continuously differentiable in $0\leq x \leq 1$. Then show that $m(\Gamma + \Gamma)> 0$ if and only if $\Gamma + \Gamma$ contains an open set, if and only if $f$ is not linear." – 2011-08-12
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0@Jonas Meyer: I've tried to use some conclusions about A+B where A an B are subsets or R^d,but any of them could help me.I just could not describe l+l in a intrinsic and proper way.So,no thoughts on the problem.I will try as many methods as I can to solve it. – 2011-08-12
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0@Jonas Meyer:some problems marked with an asterisk can find answers from other books or sb's blog,of course I uaually could not work them out by myself.This one has baffled me some time. – 2011-08-12
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2@user11211: Thanks. It might as well be stated here, before anyone posts something more serious, that containing an open set implies having positive outer measure (because open sets have positive measure). But it is stated with measure, not outer measure, so you need to know that the set is measurable. In fact, the sum of two compact subsets of $\mathbb R^2$ is compact, hence closed, hence measurable. Next, that having positive measure implies not being linear is straightforward (hint: contrapositive). Proving that if $f$ is not linear then the set contains an open set is the "hard" part. – 2011-08-12
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0@Jonas Meyer:Thanks.I can figure out some basic results.For example,C+C is measurable,sqrt(m(C+C))>=sqrt(m(C))+sqrt(m(C)).Also, I use coordinate transfer to describe C+C as {(x,f(y)+f(x-y))},and have wrote down a proof ,but not using the condition f is twice continuously differentiable.I think the way I fount out is too easy to believe the problem marked with an asterisk.Besides,I not familiar with Latex,so can not express where I stuff.I also have another idea,use Talor series on f,since f'' is not const 0. – 2011-08-12
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0@Jonas Meyer:Thank you most sincerely. You've helped me a lot.I'm now learning Latex and English(gre).May be in a few days I can write down something intresting and meaningful.I happen to know the website and really like it. – 2011-08-12