3
$\begingroup$

I've been trying to prove the following theorem for several hours with no result so far.

Claim. Let $f:\mathbb{R} \to \mathbb{R}$ be a semi-convex function, i.e. there exists a constant $C > 0$ such that $f(x) + C x^2$ is convex. Let $\operatorname{epi}(f) := \{(x,y)\in\mathbb R^2;\ f(x)\leq y \}$ denote the epigraph of $f$. Then the reach of $A = \operatorname{epi}(f)$ is positive, i.e. there exists $r > 0$ such that, for any point $(x,y)$ at distance less than $r$ to $A$, there exists a unique point in $A$ nearest to $(x,y)$.

I'll be really grateful for any help!

Thank you.

$\endgroup$
4
  • $\begingroup$ What is the reach of a set? $\endgroup$ Jan 3, 2014 at 13:08
  • $\begingroup$ The original Federer's definition: link page 15 (p 432) $\endgroup$
    – John6
    Jan 3, 2014 at 14:17
  • 1
    $\begingroup$ Perhaps you could provide the definition of reach in the question. $\endgroup$
    – Deane Yang
    Jan 3, 2014 at 17:03
  • $\begingroup$ If you can reduce the problem to the case (by an adequate change of coordinates?) where $f(0)=0$, $f$ is even and there is a sequence $(0,-\epsilon_n)$ with two closest points (therefore strictly below the horizontal axis on the graph), you are done, I think, as for any finite constant, $C$, $f(x)+Cx^2$ would not be convex in zero. $\endgroup$
    – username
    Jan 8, 2014 at 10:07

2 Answers 2

3
$\begingroup$

There is a complete discussion of this question, in general dimensions (it is true there too), in my thesis, published in Duke Math J. 1985.

$\endgroup$
1
$\begingroup$

Taking $r < \frac{1}{2C}$ should work. Indeed, if a circle of radius $r$ touches the graph of $f$ by below at two points, then the curvature $$\kappa(x) = \frac{f''(x)}{(1+(f'(x))^2)^{3/2}}$$ is larger than $-\frac{1}{r}$ at the touching points. Since $f$ lies above the circle, $\kappa \leq -\frac{1}{r}$ somewhere in between the touching points, hence $f'' \leq -\frac{1}{r}$ there. Semiconvexity completes the proof.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.