supremum of an array of a convex functions

Question

If $\{J_n\}$ is an array of a convex functions on a convex set $U$ and $G(u)=\sup J_i(u), u\in U$, how to show that $G(u)$ is convex too?

I've done this, but I am not sure about properties of a supremum.

Since $U$ is convex, $\alpha x +(1-\alpha) y)\in U$ for all $x,y\in U$.
If $G$ is convex, than it would be $G(\alpha x +(1-\alpha) y)\leq \alpha G(x)+(1-\alpha) G(y)$ i.e. $\sup J_i(\alpha x +(1-\alpha) y)\leq \alpha \sup J_i(x)+(1-\alpha) \sup J_i(y)$.

So, I've done this

$G(\alpha x +(1-\alpha) y)=\sup J_i(\alpha x +(1-\alpha) y)\leq sup \{\alpha J_i(x)+(1-\alpha)J_i(y)\}\leq \alpha \sup J_i(x)+(1-\alpha) \sup J_i(y)=\alpha G(x)+(1-\alpha) G(y)$.

Is this ok?

Answer 1

The inequalities you wrote are correct. But when you are not sure about properties of supremum, it is advisable to write down exactly why each inequality is true; this is something you have not done.

Also, an introduction of $\epsilon>0$ can reduce the level of abstraction in the argument:

1. There exists $j$ such that $G(\alpha x +(1-\alpha) y)\le J_i(\alpha x +(1-\alpha) y)+\epsilon$, by definition of supremum
2. $J_i(\alpha x +(1-\alpha) y)\leq \alpha J_i(x)+(1-\alpha)J_i(y) $, because $J_i$ is convex
3. $J_i(x)\le G(x)$ and $J_i(y)\le G(y)$, by definition of supremum
4. $G(\alpha x +(1-\alpha) y)\le \alpha G(x)+(1-\alpha) G(y)+\epsilon$, by combining 1-2-3.

It remains to use the eternally useful fact: if $a$ and $b$ are two numbers (independent of $\epsilon$) such that $a\le b+\epsilon$ for every $\epsilon>0$, then $a\le b$. "It's enough to get within an epsilon of the goal."