Let $g$ represent a mapping from $\mathbb{R}$ to $\mathbb{R}$, now consider the following function:
$\hat{g}(x) = \inf \{ w | (x, w) \in Conv(epi(g)) \}$
Where $Conv(epi(g))$ is the Convex Hull of the Epigraph of $g$
Prove that $\hat{g}(x) \leq g(x)$ for any $x$.
- Drawing a graph has definitely helped me to visualise this better, and it seems to be obvious - But I am unable to prove it formally.
My method to try and prove this is as follows: We take any arbitrary but fixed $x$, then by definition $\hat{g}(x) = w^{*} \leq w$ for all $w$ such that $(x,w) \in Conv(epi(g))$
Now we can consider 2 cases, $(x, w) \in epi(g)$ or $(x,w) \not\in epi(g)$
Case 1: $(x, w) \in epi(g)$
Then by definition, we must have $w \geq g(x)$, but at the same time, we also have $\hat{g}(x) \leq w$, and since $\hat{g}(x)$ is the greatest lower bound for $w$, we must have the case where $\hat{g}(x) \leq g(x)$
Case 2: $(x, w) \not\in epi(g)$
Then we have $w < g(x)$ and since $\hat{g}(x) = w^{*} \leq w$, then it is clear that $\hat{g}(x) < g(x)$
Hence we conclude that $\hat{g}(x) \leq g(x)$
For Case 1, I feel that it should be an equality, since this implies that $epi(g) = Conv(epi(g))$ and this only occurs when $g$ is convex (Such as $g(x) = x^{2}$), would anyone be able to guide me on how to make Case 1 more proper?
If my method of proving is wrong, appreciate it if anyone else can provide me with a better direction to start with, thanks in advance!
It is simpler: $(x, g(x)) \in epi(g) \subset Conv(epi(g))$, so that $$ \{ (x, g(x) \} \subset Conv(epi(g)) $$ and therefore $$ \begin{align} \hat{g}(x) &= \inf \{ w | (x, w) \in Conv(epi(g)) \} \\ &\le \inf \{ w | (x, w) \in \{ (x, g(x) \}) \} \\ &= g(x) \, . \end{align} $$ In other words, the set $\{ w | (x, w) \in Conv(epi(g)) \}$ contains $g(x)$, so that its infimum is $\le g(x)$.