Let $f : \mathbb{R}^M\to \mathbb{R} $
$f$ is convex if and only if $f =\sup\{\varphi:\mathbb{R}^M\to \mathbb{R}: \varphi \le f, \varphi \text { affine} \}$
To prove ($\implies$) I have to follow the following steps
(1) Fix $x \in \mathbb{R}^M$ , and $t \in \mathbb{R}$ with $t < f (x)$ We claim that there exists an affine function $\varphi : \mathbb{R}^M \to \mathbb{R}$ with $\varphi \le f$ such that $\varphi(x) > t$;
(2) Use (1), to conclude that $f =\sup\{\varphi:\mathbb{R}^M\to \mathbb{R}: \varphi \le f, \varphi \text { affine} \}:=\sup S$ by proving $f \le \sup S$ and $ \sup S \le f$
Supposing (1) is already proven, How would I prove $ (2)$?
My try:
($f \le \sup S$) :
No clue
($ \sup S \le f$): In two ways:
a) Fix $x \in \mathbb{R}^M , \forall \varphi \in S :$ $ \varphi (x)\le f(x) \implies$ Taking the sup over all such $\varphi:$ $\sup S \le f(x)$
b)By definition of sup, $\forall \varepsilon>0, \exists \varphi_\varepsilon \in S, s.t.: \sup S-\varepsilon <\varphi_\varepsilon$ and by definition of $S$, $\varphi_\varepsilon \le f$, so $\sup S-\varepsilon <f$ . Taking $\varepsilon \to 0: \sup S \le f$
What do you think? I am unsure whether or not this is correct, because I am actually not using (1), so there must be something wrong here. What is it? and How would you proceed? I don't see the point of using the inequalities involving t as stated in (1)
Assume $(1)$, and let $x\in \Bbb R^n$ and $t$ be such that $t < f(x)$. By $(1)$, there exists $\varphi$ affine such that $t < \varphi(x)$ and $\varphi \leqslant f$. It follows that $$ t < \varphi(x) \leqslant \sup_{\phi \in S} \phi(x) = (\sup S)(x). $$ Hence, we have shown that $$ t < (\sup S)(x). $$ This being true for all $t < f(x)$, one concludes that $f(x) \leqslant (\sup S)(x)$. Finally, this being true for all $x\in \Bbb R^n$, one has $f \leqslant \sup S$.