Consider the cone of continuously twice differentiable functions mapping positive reals to itself (i.e., $f\in C^2(\mathbb R_{++})$ and $f\colon \mathbb R_{++}\to\mathbb R_{++}$) that satisfy

\begin{align} f(x)&\geq 0\\ f(x) - xf'(x)- x^2f''(x)&\geq 0 \end{align} for all $x\in\mathbb R_{++}$. Let us observe that this is indeed a convex cone in that scaling and adding two functions together would preserve these two inequalities.

My question is whether it would be possible/feasible to find the extreme rays (also called generators, I believe) of this convex cone.

One can observe that each power function $f(x) = x^p$ for $p\in[-1,1]$ is in this cone, as they are positive valued and satisfy the differential condition. However it is not clear at all that these functions span this convex cone. Should one expect to have an explicit set of extreme rays?

  • $\begingroup$ $\mathbb{R}_+$ is defined as $[0,+\infty\mathclose[$ or $\mathopen]0,+\infty\mathclose[$? $\endgroup$
    – YCor
    Sep 10, 2016 at 7:38
  • $\begingroup$ A very good point. I think 0 needs to be excluded, so let me replace all occurrences of $\mathbb R_+$ with $\mathbb R_{++}$, which I believe is used for zero-excluded set. $\endgroup$
    – JLehec
    Sep 10, 2016 at 7:48

1 Answer 1


You are looking at the intersection $C = C_1 \cap C_2$ of two cones, $C_1$ defined by the condition $f(x) \ge 0$ and $C_2$ defined by the condition $Df(x) \ge 0$, where $D=(1+x\partial_x)(1-x\partial_x)$. In a sense, $C_2 = D^{-1} C_1$, where the right-hand-side is interpreted as the pre-image of $C_1$ with respect to $D$.

Now, $C_1$ does not have extreme rays (those that do not belong to the sum of any two different rays from the same cone), because any non-negative function with non-trivial support can always be written as a convex linear combination with positive coefficients of two other non-negative functions with smaller supports. If you embed $C^2(\mathbb{R}_{++})$ in distributions, then then the closure of $C_1$ does have extreme rays, which are the Dirac $\delta$ distributions, $\delta(x-y)$.

Next, look at $C_2$. Using an explicit right-inverse $D^{-1}_0$ of the operator $D$, we can write $C_2 = \ker D + D^{-1}_0 C_1$, where $\ker D = \mathbb{R}_+ x + \mathbb{R}_+ x^{-1}$. A right inverse can be given, if I got all the calculations right (edit: turns out that I did miss an overall minus sign in the previous version), by the integral operator \begin{multline*} D^{-1}_0[g](x) = -\int_0^x \frac{(x-y)(x+y)}{2xy^2} \chi_0(y) g(y) \, dy \\ - \int_x^\infty \frac{(y-x)(x+y)}{2xy^2} \chi_\infty(y) g(y) \, dy , \end{multline*} where $\chi_0(y)$ and $\chi_\infty(y)$ are any two non-negative continuous functions such that $\chi_0(y) + \chi_\infty(y) = 1$, with $\chi_0(y)$ zero on a neighborhood of $0$ and $\chi_\infty(y)$ zero on a neighborhood of infinity. The way it's written, the integrals are well-defined for any continuous $g(x)$. Note that, because the integral kernel is non-positive, $g(x) \ge 0$ implies that $D^{-1}_0[g](x) \le 0$. Note that, because the integral kernel is non-negative, $g(x) \ge 0$ implies that $D^{-1}_0[g](x) \ge 0$.

Finally, with the above observations, we can conclude that $$ C = C_1 \cap (\ker D + D^{-1}_0 C_1) = D^{-1}_0 C_1 + \ker_{\ge} D , $$ where $\ker_{\ge} D = \mathbb{R}_+ x + \mathbb{R}_+ x^{-1}$. Now, $D^{-1}C_1$ does not have any extreme rays either, by the same argument as for $C_1$. On the other hand, $\ker_{\ge} D$ does have extreme rays, those generated by $x$ and $x^{-1}$. Moreover, since $\ker_{\ge} D \cap D^{-1}_0 C_1 = \varnothing$, these two rays remain extreme when we take the sum of these two cones to get $C$.

So, it seems that the conclusion is that the cone $C$ has only the two extreme rays generated by $x$ and $x^{-1}$ and no other ones. Clearly, $C$ is very far from being even the closure of the convex linear combinations of its extreme rays.

  • $\begingroup$ Thank you very much. It will take me some time to reconstruct the inverse differential equation you gave, but on a first look I am confused that $D^{-1}[g]$ is nonnegative. For $f(x)=-x^{1+\epsilon}$, $Df$ should be in $C_2$, if I'm not miscalculating. $\endgroup$
    – JLehec
    Sep 11, 2016 at 0:19
  • 1
    $\begingroup$ Hmm, I may be off by an overall minus sign somewhere... $\endgroup$ Sep 11, 2016 at 7:51
  • 1
    $\begingroup$ Right, I was missing an overall minus sign in my $D^{-1}_0$ (fixed now). Unfortunately, that makes the discussion a bit more complicated. To get an explicit description of $C$ one now has to understand the set of those $g$ for which $D^{-1}_0[g]$, though negative, grows no faster than $O(x)$ at infinity and $O(x^{-1})$ at zero. $\endgroup$ Sep 11, 2016 at 9:31

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.