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I am trying to understand a step in this proof.

Let $\{f_j\}_{j \in J}$ be an arbitrary family of convex functions: $f_j: X \to \mathbb{R}$ where $X \subseteq \mathbb{R}^n $ is convex. Show that $f(x):= \sup\{f_j(x)| \; j\in J\}$ is convex.

Proof:

$$f(x) := \sup \{ f_j(x) | \; j \in J\}$$

$$f(\lambda x + (1-\lambda)y) := \sup \{f_j(\lambda x + (1-\lambda)y | \; j \in J\}$$

$$\sup \{f_j(\lambda x + (1-\lambda)y) | \; j \in J\} \leq \sup \{\lambda f_j(x) + (1-\lambda)f_j(y)| j \in J\} $$ $$\leq \sup\{\lambda f_j(x)|j\in J\}+\sup\{(1-\lambda)f_j(y)|j\in J\} $$

The first inequality seems to follow from the convexity of each $f_j$ - is this correct? But then where does the last inequality follow from, where the $\sup(\cdot)$'s are can be split apart? Should it not just be an equality? Any help is very much appreciated.

codenoob
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    Correct on the first inequality. The second is an inequality because if you maximise both of them individually you are able to pick a different $j$ for each one, thus achieving a higher value. – fGDu94 Dec 12 '19 at 19:55
  • Sorry, i am struggling to think of an example in my head, could you provide one? if we have sup(3+1) and sup(3) + sup(1) we get an equality. Same with sup(-1+3) = sup(-1) + sup(3). Is it because I am only considering numbers, when I should be considering functions that may have asymptotic properties? – codenoob Dec 12 '19 at 20:03
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    You need to consider sets of functions. For example, suppose you only have two functions and $f_1(x)=1, f_2(x)=0, f_1(y)=0, f_2(y)=1$. Then $\sup(f_i(x)+f_i(y) | i \in {1,2}) = 1$ but $\sup(f_i(x) | i \in {1,2}) + \sup(f_i(y) | i \in {1,2}) = 2$ – user6247850 Dec 12 '19 at 20:08
  • Thank you! Makes sense now. – codenoob Dec 12 '19 at 20:14
  • exactly what user6247850 said – fGDu94 Dec 12 '19 at 20:33

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