4
$\begingroup$

The complete claim is,

Let $F$ be an ordered field and $\bar F=F\bigcup\{+\infty,-\infty\}$. Let $f:A \to \bar F$ be a convex map from affine space $(A,V)$ to affine space $(F,F)$, then $f(C)$ is a convex set if $C$ is convex.

Here $F$ can be treated as $\Bbb R$ except for that we do not use its topology and metric. $A$ is a point set, and $V$ is a vector space.

A convex function is defined using epigraph,

$f$ is a convex set if its epigraph $\text{epi} f=\{(x,y)\in A\times F:y\ge f(x)\}$ is a convex set.

I am not sure if this claim is right but it seems so. I find difficulty in proving it. The following is an attempt, but no contradiction is found.

If $f(C)$ is not convex, then there exists $y_1,y_2\in f(C)$ st. $y_1>y_2$ and $\theta y_1 + (1-\theta) y_2 \notin f(C)$ for some $\theta \in (0,1)$. Let $y_3=\theta y_1 + (1-\theta) y_2$, and suppose $f(x_1)=y_1, f(x_2) = y_2$ for some $x_1,x_2\in C$. Let $x_3 =\theta x_1 + (1-\theta) x_2)$, then $(x_1,y_1),(x_2,y_2)\in \text{epi} f \Rightarrow (\theta x_1 + (1-\theta) x_2, \theta y_1 + (1-\theta) y_2) = (x_3,y_3) \in \text{epi} f \Rightarrow y_3 > f(x_3)$

The last inequality $y_3>x_3$ does not contradict with Jensen's inequality.

I need help with this. If it is not comfortable to deal with affine space, we might just assume $f:\Bbb R^n \to \Bbb R$, and just try not to use topology information (such like continuity) and metric in the proof. I add "real analysis" as a tag of this question for this reason.

Again, note the claim might not be true. If it is not true, hope someone could come up with a counterexample.

Thank you!

$\endgroup$
  • 1
    $\begingroup$ The characteristic function $\chi_C$ of any convex set $C$ is convex and has image $\{0,\infty\}$, defined by $\chi_C(x) = 0$ for $x\in C$ and $=\infty$ otherwise. $\endgroup$ – user251257 Sep 1 '16 at 22:51
  • $\begingroup$ @Tom in the question, $f$ is supposed to be scalar valued. $\endgroup$ – user251257 Sep 2 '16 at 5:57
  • $\begingroup$ @Tom the epigraph is a convex set, not the graph. $\endgroup$ – Michael Grant Sep 2 '16 at 15:38
  • $\begingroup$ @MichaelGrant the question isn't about the epigraph, but the graph. $\endgroup$ – user251257 Sep 2 '16 at 16:58
  • $\begingroup$ Ah, then really, I should be directing that comment to the OP :-) $\endgroup$ – Michael Grant Sep 2 '16 at 16:58
1
$\begingroup$

At least for $F=\mathbb R$ and $A=V=\mathbb R^n$ it is pretty simple.

If you allow $\infty$, then the statement is not true in general. The characteristic function $f=\chi_C$ of any convex set $C\subseteq A$, defined by $$f(x) = \begin{cases} 0, & x\in C, \\ \infty, & x\notin C, \end{cases}$$ is convex and has image $f(A)=\{0,\infty\}$. Obviously the image isn't convex.

If you only allow finite values, then it is true. On a finite dimensional space, a convex function $f$ is continuous on the interior of its proper domain. Since $\mathbb R^n$ is open, the image $f(C)$ must be connected for any connected $C\subseteq A$. Notice that convex sets are also connected. Now, the connected subsets of $\mathbb R$ are the intervals and thus are convex.

$\endgroup$

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.