Let $\Omega$ be a bounded convex domain in $\mathbb{R}^n$ and $u$ be a continuous convex function on the closure $\overline{\Omega}$. Given a boundary point $p\in\partial\Omega$, $u$ is said to have infinite slop at $p$ if $$\lim_{t\rightarrow0^+}\frac{u((1-t)p+tq)-u(p)}{t}=-\infty$$ for some $q\in\Omega$ (which implies the same limit for every $q\in\Omega$). Otherwise, $u$ is said to have finite slop at $p$.

On the other hand, the Legrendre transform domain of $u$ is the subset of the dual vector space $\mathbb{R}^{*n}$ where the Legendre transform of $u$ is defined (i.e. takes finite values). If $u$ is $C^1$ in $\Omega$, this domain is nothing but $\{Du(q)\mid q\in\Omega\}$.

The following question has an obvious positive answer when $n=1$, but I had some trouble trying to prove the "if" part in general:

Question. Is it true that for any convex function $u\in C^0(\overline{\Omega})$ with $u|_{\partial\Omega}=0$, the Legendre transform domain of $u$ is bounded if and only if $u$ has finite slope at every boundary point?

Is there a literature addressing this issue?

  • The basic example I'm aware of is $u(x)=(1-x^2)^{-1}$ for $|x|<1$, where $u^*$ has unbounded domain. Your hypothesis $u|_{\partial \Omega}=0$ is pretty strong. Don't we know Claim 1: $dom(u^*)$ equals the closure of the convex hull of $\{\partial u(y),~~y\in \Omega\}$ ? And Claim 2: if $u$ has finite slope at every boundary point, then the set $\{\partial u(y),~~y\in \Omega\}$ of normals should be bounded. That would imply $dom(u^*)$ is bounded, if the two claims are correct. – J. Martel Nov 7 at 12:50
  • @J. Martel Actually the domain I'm considering is exactly $\{\partial u(y), y\in\Omega\}$, and your claim 2 is the essence of the problem: the claim can fail if for some $p\in\partial\Omega$ where $u$ has finite slope, the slope of $u$ along an inward pointing vector tends to $-\infty$ when the vector approaches a tangent vector of $\partial\Omega$ at $p$. I could not rule out this situation or find examples with $u|_{\partial\Omega}=0$. – Xin Nie Nov 8 at 6:18
  • I guess I don't understand the meaning of ``slope". I figured it was basically the directional derivative $\langle v, n \rangle$ of normal vectors $n$ of the graph of $u$, for unit vectors $v$. And i figured that if $u$ has bounded values over the domain (which i expect is consequence of $u|_\Omega =0$), then all the normals are bounded, and therefore all directional derivatives. But maybe I'm wrong here... – J. Martel Nov 9 at 0:44
  • @J. Martel The "slope" is basically the normal vector $n$ itself rather than its derivative. $\text{dom}(u^*)$ being unbounded means that $n(x_n)$ tends to be "horizontal" for a sequence $x_n\in\Omega$ going to the boundary, or equivalently, the tangent plane to the graph of $u$ at $x_n$ tends to be ''vertical". – Xin Nie Nov 9 at 7:27
  • Then isn't the question tautological? The domain of $u^*$ is the set of slopes. – J. Martel Nov 9 at 17:32

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