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.

Improve this question
$\endgroup$
4
  • $\begingroup$ What is the reach of a set? $\endgroup$
    – Alex Degtyarev
    Commented Jan 3, 2014 at 13:08
  • $\begingroup$ The original Federer's definition: link page 15 (p 432) $\endgroup$
    – John6
    Commented Jan 3, 2014 at 14:17
  • 1
    $\begingroup$ Perhaps you could provide the definition of reach in the question. $\endgroup$
    – Deane Yang
    Commented 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
    Commented Jan 8, 2014 at 10:07

2 Answers 2

Reset to default
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.

Improve this answer
$\endgroup$
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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .