Why is pointwise maximum a convex function?

Asked Apr 10 '14 at 01:23

Active Apr 10 '14 at 01:37

Viewed 2,318 times

It seems like if you have a family of function

$$g = \{a(x), \: b(x), \: c(x), \:d(x)\}$$

$$\text{given} \:\: f(x):= max(g),$$

$$\text{if} \: f(1) = a(1), \: f(2) = b(2), \: f(3) = c(3), \: f(4) = d(4) $$

$f\:$ have a possibility of not being continous

Thus the epigraph is not a convex set, am I misunderstanding something?

Edit: I am assuming $a, b, c, d$ to be all convex, but they have different domains that don't overlap

edited Apr 10 '14 at 01:37

asked Apr 10 '14 at 01:23

user2654176

Which assumptions do you have about the convexity and/or continuity of $a, b, c, d$? – hmakholm left over Monica Apr 10 '14 at 01:28
I am assuming a, b, c, d to be all convex, however 'a' might have different domain than 'b' – user2654176 Apr 10 '14 at 01:35
By convention, a convex function is $+\infty$ outside its domain. (This is to ensure that the maximum of convex functions is convex.) – p.s. Apr 10 '14 at 02:24
The epigraph of the $\max$ is the intersection of the epigraphs, hence $f$ must be convex (and continuous on the relative interior of its domain). – copper.hat Apr 10 '14 at 04:57
Thanks for all the comments, I think I get the intuition now, thanks – user2654176 Apr 10 '14 at 22:24
See related question http://math.stackexchange.com/questions/402919/pointwise-supremum-of-a-convex-function-collection – Shlomi A Jan 19 '17 at 08:10

0 Answers0