Unions of convex functions

Question

Let $A$ and $B$ denote open convex subsets of $\mathbb{R}^n$. Suppose $f$ and $g$ are convex functions on $A$ and $B$ respectively, mapping into the real line. It seems likely that if $f$ and $g$ agree on $A \cap B$, and if $A \cup B$ is convex, then $f \cup g : A \cup B \rightarrow \mathbb{R}$ is a convex function.

Question. Is this true?

The assumption that $A$ and $B$ are open is necessary to prevent the following counterexample: $n=1$, $A = [0,1)$, $B=[1,2]$, and $f$ and $g$ are constant functions of different heights.

@HKLee, if this is necessary for the proof to go through, then I guess so :) do you have a counterexample in mind? — goblin GONE, Apr 06 '17 at 11:05
If we permit that $f,\ g$ may be unbounded, then $f=0$ on $A-A\cap B$, $f=g=\infty$ on $A\cap B$ and $g=0$ on $B-A\cap B$. — HK Lee, Apr 06 '17 at 11:07
@HKLee, I don't think $f$ would be convex though, because certain lines would dip below the graph of $f$ on $A \cap B$. — goblin GONE, Apr 06 '17 at 11:09
If $f$ and $g$ are finite on $A \cap B$, the assertion should hold. It is easy to see that it is sufficient to consider only the case $n = 1$. And for $n = 1$, the statement seems to be easy to prove. — gerw, Apr 06 '17 at 11:16

score 0 · Answer 1 · answered Apr 06 '17 at 11:24

0

I will follow gerw's comment :

If $f :(a,c)\rightarrow \mathbb{R},\ g:(b,d)\rightarrow \mathbb{R}$ are convex and $f=g$ on $(b,c)$, then consider $x<b,\ y>c$ Then consider convex hull of $$(x,f(x)),\ (b,f(b)),\ (c,g(c)),\ (y,g(y))$$

answered Apr 06 '17 at 11:24

HK Lee

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Unions of convex functions

1 Answers1