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).
[Math] Real analysis by E. M. Stein: Chapter 1,Problem 7
real-analysis
Best Answer
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.
EDIT:
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 $(x,z)$.