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So I've seen a few different definitions of quasiconvexity of a function and I cannot, after a bit of working, figure out how all of them are related:

Let $X$ be a convex subset of a real vector space:

  1. $\dagger$ A function $f : X \to \mathbb{R}$ is quasiconvex if $f(\lambda x + (1-\lambda)y) \le \max \{ f(x) , f(y) \}$.
  2. $\dagger$ A function $f : X \to \mathbb{R}$ is quasiconvex if $f^{-1} ((-\infty, \alpha))$ is convex $\forall \alpha \in \mathbb{R}$
  3. $\dagger$ A function $f : X \to \mathbb{R}$ is quasiconvex if $f^{-1} ((-\infty, \alpha])$ is convex $\forall \alpha \in \mathbb{R}$
  4. $\dagger$ A function $f : \mathbb{R}^{mn} \to \mathbb{R}$ that is borel regular and locally integrable is quasiconvex if $$ f(A) \le \frac{1}{\mu(D)} \int_D f(A + \nabla \phi(x)) \, dx \; \forall A \in \mathbb{R}^{mn} \; \forall D \text{ bounded} \subset \mathbb{R}^n \; \forall \phi \in W_0^{1,\infty}(D; \mathbb{R}^m) $$

where $\dagger$ is the source of the statement.

I understand how $1$ and $3$ are connected.

I feel as though I should be able to say something about the connection between $2$ and $3$ (i.e. if $f$ is $1$ then it's possible that adding $\alpha$ opens up a whole in the preimage meaning that $x$ so that $f(x) = \alpha$ is somewhere arbitrarily far away, but then adding the preimage of a right hand neighborhood of $\alpha$ fills this hole). Also there is a definition of strict quasiconvexity which is the same definition of $1$ but with $\le$ replaced with $<$ and I believe I see how this is similar to $2$. If this is the case then I understand that $2$ is just the strict case and $3$ is the not-strict case.

As for $1,2,3$ connected to $4$ at all I'm unsure.

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Concept 4 is very different from 1,2,3. It is just a different generalization of convexity which happens to have the same name (which, incidentally, makes for some ambiguous questions posted on this site). Since the definition 4 is due to Morrey, it is a good idea to call it Morrey quasiconvexity to avoid confusing with 1,2,3.

Properties 1 and 3 are equivalent, as you understood. So are 2 and 3:

$3\implies 2$ because $\{f<a\}$ is the union of $\{f\le a-1/k\}$ over $k$. The union of an increasing family of convex sets is convex.

$2\implies 3$ because $\{f\le a\}$ is the intersection of $\{f< a+1/k\}$ over $k$. The intersection of any family of convex sets is convex.

To reformulate strict quasiconvexity (as in item 1, but with $<$) in terms of level sets would not be easy. The natural candidate is: the sets $\{f\le a\}$ are strictly convex... but it does not work. For example, $f(x,y)=\max(x^2+y^2,1)$ is a quasiconvex function that is not strictly quasiconvex. Yet, all sets of the form $\{f<a\}$ or $\{f\le a\}$ are round disks.

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  • $\begingroup$ Thanks, do you have a reference for the union of an increasing family of convex sets is convex? $\endgroup$ – DanZimm Jul 15 '14 at 5:40
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    $\begingroup$ Oop my apologies, I think the proof would go something like if $x,y \in \cup A_i$ where $A_i$ is an increasing family then at some $n$ $x,y \in A_n$ and $A_n$ is convex so it follows $\endgroup$ – DanZimm Jul 15 '14 at 5:44
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    $\begingroup$ @DanZimm Yes, that's how it goes. $\endgroup$ – user147263 Jul 15 '14 at 5:48
  • $\begingroup$ Please help me in the following question: math.stackexchange.com/questions/1220706/… $\endgroup$ – Blind Apr 5 '15 at 17:29

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