1
$\begingroup$

A piecewise function $F(x_1,x_2)$ is continuous, so are its 1st partial derivatives. However, its 2nd partial derivatives are discontinuous, that is , $F$ is not of class $C^2$. But its Hessian satisfies the convexity condition piecewise (even turns out to be symmetric), i.e. all its principal minors are non-negative, so is $F$ jointly convex even it is not of class $C^2$?

$\endgroup$
2
  • $\begingroup$ Do the second order derivatives exist at every point of $\mathbb R^2$, even at the boundaries of "pieces"? If not, how smooth are the pieces? $\endgroup$ Jun 1, 2013 at 13:29
  • $\begingroup$ Specifically, $F''_{11}$ exists everywhere except for lines $x_1=S$ and $x_1+x_2=S$, $F''_{12}$ and $F''_{22}$ exist everywhere except for line $x_1+x_2=S$,where $S$ is a positive constant. Moreover, all $F''_{ij}$s are "piecewise" nonnegative, and the function is "piecewise" convex. The specific function seems convex to me, but I just wonder if there is a general conclusion, when is the "class of $C^2$" condition unnecessary? $\endgroup$
    – jorter.ji
    Jun 2, 2013 at 16:54

1 Answer 1

2
$\begingroup$

Proposition. Suppose $U\subset \mathbb R^n$ is a convex domain, and $F:U\to\mathbb R $ is a $C^1$ function. Suppose further that $F$ is second-order differentiable on $U\setminus N$, where $N$ is a closed set that is locally $(n-1)$-rectifiable. If the Hessian matrix of $F$ is positive semidefinite on $U\setminus N$, then $F$ is convex.

(Remark: a set that is a finite union of smooth $(n-1)$ hypersurfaces is locally $(n-1)$-rectifiable.)

Proof. A $C^1$ function $F:U \to \mathbb R$ is convex if and only if its derivative is monotone: $$\langle \nabla F(a)-\nabla F(b),a-b\rangle \ge 0,\quad \forall a,b\in U \tag1$$ For a given pair $a,b$, the intersection of the line segment $[a,b]$ with $N$ may happen to be infinite. But recall that the piece of $N$ in a ball surrounding $[a,b]$ has finite $(n-1)$-dimensional measure. Fubini's theorem implies that for almost every vector $u$ orthogonal to $[a,b]$, the shifted segment $[a+u,b+u]$ meets $N$ in a finite set (possibly empty). By continuity, it suffices to prove (1) for a dense subset of pairs $a,b$. Thus, we may assume that $[a,b]\cap N$ is finite.

The restriction of $F$ to $[a,b]$, denoted $f(t)=F((1-t)a+tb)$, has nonnegative second derivative outside of a finite subset of $[0,1]$. Since $f'$ is continuous on $[0,1]$, it follows that $f'(1)\ge f'(0)$. In terms of $F$ this says precisely that (1) holds. $\Box$

$\endgroup$
6
  • $\begingroup$ Thanks, but I am a little confused with your link. (1) A $C^1$ function $F$ is convex $\iff$ its derivative is monotone or $\iff$ $\nabla F$ is monotone? But $\nabla F$ is gradient, not derivative. (2) The (total?) derivative of $F(x+tv)$ is $\langle\nabla F(x+tv), v\rangle$? $\langle,\rangle$ means inner product? Moreover, the Fubini's theorem from Wikipedia is "a result which gives conditions under which it is possible to compute a double integral using iterated integrals", it seems to me different from the one you used? $\endgroup$
    – jorter.ji
    Jun 3, 2013 at 23:07
  • 1
    $\begingroup$ @jorter.ji (1) I'm in the habit of calling $\nabla F$ the derivative of $F$. You can call it the gradient if you prefer. (2) By $F(x+tv)$ I mean the function of one real variable $t$, which is the restriction of $F$ to a line. Its derivative is computed using the chain rule. Yes, $\langle,\rangle$ is inner product. (3) Not really Fubini's theorem, it's called the slicing lemma in geometric measure theory. See page 7 here for example. $\endgroup$ Jun 3, 2013 at 23:31
  • $\begingroup$ Thanks. (1) and (2) now make more sense to me. But I am still lost on (3). The slicing lemma in the pdf link seems not very closely related, is there a "original" (Fubini's) lemma/theorem that leads to the implication in your answer? $\endgroup$
    – jorter.ji
    Jun 4, 2013 at 0:01
  • 1
    $\begingroup$ @jorter.ji It is related, but the statement I really meant is: the $d$-dimensional measure of a set $A\subset \mathbb R^n$ is the integral of $(d-k)$-dimensional measure of intersections of $A$ with by parallel $(n-k)$-dimensional hyperplanes. In this case let $d=k=n-1$. The parallel hyperplanes become lines parallel to $[a,b]$. Also, $d-k=0$ and the $0$-dimensional measure is just the number of points. So, if the number of points in the intersection is infinite for a set of positive measure, than the integral is infinite, contradicting the finiteness of $(n-1)$-dimensional measure. $\endgroup$ Jun 4, 2013 at 0:13
  • $\begingroup$ @jorter.ji A standard reference is: P. Mattila, "Geometry of sets and measures in Euclidean spaces. Fractals and rectifiability". Cambridge Studies in Advanced Mathematics, 44. Cambridge University Press, Cambridge, 1995. $\endgroup$ Jun 4, 2013 at 0:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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