4

In Hoang Tuy, Convex Analysis and Global Optimization, Kluwer, pag. 46, I read:

A positive combination of finitely many proper convex functions on $R^n$ is convex. The upper envelope (pointwise supremum) of an arbitrary family of convex functions is convex.

In order to prove the second claim the author sets the pointwise supremum as:

$$ f(x) = \sup \{f_i(x) \mid i \in I\} $$

Then

$$ \mathrm{epi} f = \bigcap_{i \in I} \mathrm{epi} f_i $$

As "the intersection of a family of convex sets is a convex set" the thesis follows.

The claim on the intersections raises some doubts for me.

If $(x,t)$ is in the epigraph of $f$, $f(x) \leq t)$ and, as $f(x)\geq f_i(x)$, also $f_i(x) \leq t)$; therefore $(x,t)$ is in the epigraph of every $f_i$ and the intersection proposition follows.

Now, what if, for some (not all) $\hat{i}$, $f_{\hat{i}}(x^0)$ is not defined? The sup still applies to the other $f_i$ and so, in as far as the sup is finite, $f(x^0)$ is defined. In this case when $(x^0,t) \in \mathrm{epi} f$ not $\in \mathrm{epi} f_{\hat{i}}$ too, since $x^0 \not \in \mathrm{dom}\, f_{\hat{i}}$.

So it is simple to say that $f$ is convex over $\bigcap_{i \in I} \mathrm{dom} f_i\subset \mathrm{dom} f$, because in this set every $x \in \mathrm{dom}\, f_{\hat{i}}$.

What can we say when $x \in \mathrm{dom}\, f$ but $x \not \in \mathrm{dom}\, f_{\hat{i}}$ for some $i$?

Rodrigo de Azevedo
  • 23,223
antonio
  • 421
  • The sup still applies to the other fi... No Sir, one cannot modify the definition like that, at every point the supremum takes into account every index i. – Did May 26 '13 at 14:41
  • @Did: sorry when you say "takes into account every index i", do you mean that every $f_i(x)$ has the same domain? So I cannot have a situation where, for some $x^0,i,j$, $;x^0 \in \mathrm{dom}, f_i$ and $x^0 \not\in \mathrm{dom}, f_j$. – antonio May 26 '13 at 16:02
  • @Did So, since the supremum takes into account every index $i$, HOW does the supremum take into account indexes $i$, at points such that $f_i(x)$ is not defined? Is it as the accepted answers says and we just set $f_i$ to $+\infty$ if $x\not \in Dom(f_i)$? If so, is that such a common convention (or is the $f_i$ having different domains so uncommon) that it needs not be specified in the statement? The reason I ask is that I have been thinking about what part of the statement is false in the situation being considered and I don't see it (hence, I want to (wrongly) apply the theorem). – user106860 Nov 18 '16 at 01:37
  • 1
    @user106860 Once again, the supremum applies to functions defined on a common domain $D$. If they are not, the trick is usually to extend each function by $-\infty$ outside of its original domain (rather than $+\infty$), so that they are indeed defined on a common domain. – Did Nov 18 '16 at 10:33

1 Answers1

4

I think it is either assumed that the $f_i$ are defined on the same domain $D$, or that (following a common convention) we set $f_i(x)=+\infty$ if $x \notin \mathrm{Dom}(f_i)$. You can easily check that under this convention, the extended $f_i$ still remain convex and the claim is true.

Albert
  • 9,487
  • +1. As the function is defined 'proper' it can take $+\infty$ values outside its 'effective domain'. As you observe it works in this case. Also the related sup value is different. If some $f_i$ is $+\infty$ so is the sup; otherwise it would be the sup of the $f_i$ defined at $x$. – antonio May 26 '13 at 18:59