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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).

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Sorry.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. – student Aug 12 '11 at 4:09
The 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. – Gerry Myerson Aug 12 '11 at 4:13
It 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." – Jonas Meyer Aug 12 '11 at 4:20
"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." – Jonas Meyer Aug 12 '11 at 4:21
@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. – Jonas Meyer Aug 12 '11 at 4:45

Another way to state this problem: Show that if $f(x)$ is linear, then $m(C + C) = 0$, and that if $f(x)$ is not linear, then $C + C$ contains an open set.

The set $C + C$ is $\{(x + z, f(x) + f(z): 0 \leq x \leq 1, 0 \leq y \leq 1\}$. By just the form $y = mx + b$, it should not be hard for you to show that if $f(x)$ is linear then $C + C$ is on a line whose equation you can determine.

The second part is the harder part. The idea is that if $f''(x) \neq 0$, then without loss of generality (by $C^2$ness) we can assume that $0 < x < 1$. The goal will be to show that $C + C$ contains an open set near $(2x, 2f(x))$. Note that a point $(x + z, f(x) + f(z))$ for $z$ near $x$ can be rewritten as $(w, f(x) + f(w - x))$ where $w$ is near $2x$.

Viewed as a function of $x$ now, $f(x) + f(w - x)$ has derivative zero at $x = w/2$, but it has nonvanishing second derivative at $x = w/2$, so long as $f''(w/2)$ is nonzero. Thus as $x$ varies a vertical segment with endpoint $(w/2, 2f(w/2))$ is traced out. This holds for any $w$ in an interval for which $f''(w/2) \neq 0$, and the lengths of the segment are bounded below (and are either all above or all below $(w/2, 2f(w/2))$) by the $C^2$ condition. So you get what you need.


After writing that up, another (quicker) way to do the second part occurred to me.. The Jacobian determinant of $(x,z) \rightarrow (x + z, f(x) + f(z))$ is $f'(z) - f'(x)$. This is going to be nonzero if $x \neq z$ are both near a point $y$ where $f''(y) \neq 0$. So the inverse function theorem gives that the image of this map contains an open disc centered at the image of some such $(x,z)$.

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:Thank you so much for giving the luminate answer.Your second idea using Jacobian determinant is really fascinating. – student Aug 12 '11 at 15:28

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