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Motivation

In thermodynamics, given a free-energy function $f$ of a conserved parameter $\phi$ for some material, we can predict whether that material will prefer to separate into multiple phases from convexity properties of $f$. To do this, we can construct the lower convex envelope (or Fenchel biconjugate) $\operatorname{conv} f$, and see where it is below the original function. Those values of $\phi$ are unstable.

I'm working with this type of system for two conserved parameters, and I would like to formalise some intuitions about these objects. In particular, there are three properties (listed below) that I would like a proof of.

I am not a convex analyst, and unfortunately none of these three were listed in the convex analysis textbooks I could get my hands on [1,2]. I am looking either for a reference for these properties, or counterexamples if they are false. Because this is not my field, I would appreciate it if — in case the property, as stated, is false, but becomes true with a minor change of assumptions — you could point me to the corrected property.

I've put all three properties together because it seems to be that they are related, but I will happily resubmit this as three questions if that is better for this site.

Definitions

Given a function $f: \mathbb R^n \to \mathbb R \cup \{\infty\}$, use the usual definition of its convex envelope $\operatorname{conv}f$ as the greatest convex function majorised by $f$: $$\operatorname{conv}f := \sup_h \left\{h: \mathbb R^n \to \mathbb R \cup \{\infty\} \ \big|\ h\leq f,\ h\text{ convex} \right\} \,.$$ Then, we define the instability set $\Xi_f$ of $f$ as the set on which it differs from its lower convex envelope: $$\Xi_f := \left\{\phi\in \mathbb R^n \ \big|\ f(\phi) > (\operatorname{conv}f)(\phi) \right\} \subseteq \mathbb R^n\,.$$

Given these, we can move on to the three intuitions.

Property 1 (Linear functions have no effect on phase separation.)

Two functions $f:\mathbb R^n \to \mathbb R \cup \{\infty\}$ and $g: x\mapsto f(x) + a\cdot x + b$ (where $a,b \in \mathbb R^n$, and $\cdot$ is the usual scalar product) which differ only by an affine function share sets of instability: $\Xi_f = \Xi_g \,.$

Property 2 (On the set of instability, the convex envelope has vanishing curvature.)

The curvature of $\operatorname{conv} f$ restricted to $\Xi_f$ always vanishes in at least one direction: $$\left.\det \left( \mathcal H_{\operatorname{conv}\!f} \right)\right|_{\Xi} = 0 \,,$$ where $\mathcal H$ is the hessian matrix $(\mathcal H_g (\phi))_{ij} := \partial^2 g(\phi) / \partial \phi_i \partial \phi_j$. Here I'm happy to assume whatever smoothness properties for $f$ are necessary, e.g. that it is twice continuously differentiable inside the set where it is finite.

Property 3 (Smoothness of the binodal.)

This one I'm going to state in simplified form, so that it at least works for the system I'm interested in. Of course, more general versions are welcome.

Assume $f$ is smooth on an open set $U \subset \mathbb R^2$. Then, connected components of the instability boundary $\partial \Xi_f$ are smooth curves in $U$.

References

[1] J. M. Borwein and A. S. Lewis, Convex analysis and nonlinear optimization: theory and examples. New York, NY: Springer New York, 2000. DOI: 10.1007/978-1-4757-9859-3.

[2] J. B. Hirart-Urruty and C. Lemaréchal, Convex analysis and minimization algorithms I. [Berlin]: Springer-Verlag Berlin Heidelberg, 1993. ISBN: 0-387-56850-6.

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    $\begingroup$ Welcome to MathOverflow (MO). According to MO guidelines, users should not respond to posts that "request answers to multiple questions". Since the answer to your first question is obvious and your third question has been answered, please post your second question separately. $\endgroup$ Commented May 19 at 19:23

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$\newcommand\R{\Bbb R}$Counterexample to Property 3: $f(x,y)=x^2(x-1)^2 y^2(y-1)^2$ for $(x,y)\in U:=\Bbb R^2$. Then $f$ is a smooth function, but a connected component of $\partial\Xi_f$ contains the boundary (say $B$) of the open unit square (say $S$), which is not a smooth curve.

Indeed, take any convex function $h$ such that $h\le f$. Since $f=0$ on $B$, we have $h\le0$ on $B$ and hence on $S$. So, $\operatorname{conv}f\le0$ on $S$. On the other hand, $f>0$ on $S$, so that $S\subseteq\Xi_f$. Since the constant zero function is convex and $\le f$ and $=f$ on $B$, we see that $\operatorname{conv}f=0=f$ on $B$. So, $B\cap\Xi_f=\emptyset$. So, $B=\partial S\subseteq\partial\Xi_f$.

For an illustration, here is the graph $\{(x,y,f(x,y))\colon(x,y)\in[-0.2,1.2]^2\}$:

enter image description here

Property 1 is obvious. Indeed, letting $q_{a,b}(x):=q(x)+a\cdot x+b$ for any function $q$, we see that $h_{a,b}$ is a convex function such that $h_{a,b}\le f_{a,b}$ iff $h$ is a convex function such that $h\le f$, so that $\operatorname{conv}(f_{a,b})=(\operatorname{conv}f)_{a,b}$, and hence $f_{a,b}-\operatorname{conv}f_{a,b}=f-\operatorname{conv}f$.

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  • $\begingroup$ Thank you for the answer, it's very clear and the diagram really helps — I can't believe I missed that family of counterexamples. I would never have been able to come up with the 'obvious' proof of property 1, but I think I can follow it. To verify, the explanation behind $\operatorname{conv}(f_{a,b}) = (\operatorname{conv}f)_{a,b}$ is that $\sup\{h|h\leq f\} = \sup\{h|h_{a,b}\leq f_{a,b}\} = \sup\{h_{a,b}|h_{a,b}\leq f_{a,b}\} - a\cdot \mathrm{id} - b$, is that right? $\endgroup$ Commented May 20 at 16:35
  • $\begingroup$ @confusedandbemused : Concerning Property 1 -- yes, I think you can express it this way. $\endgroup$ Commented May 20 at 16:39

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