Let $f : \mathbb{R} \to \mathbb{R}$ be a convex function, and define the Legendre transform of $f$ by $$L(a) := \sup_{x \in \mathbb{R}} (ax - f(x)) \ .$$ I wish to show that $$f(a) = \sup_{x \in \mathbb{R}} (ax - L(x)) \ .$$ Let $x \in \mathbb{R}$ and we note that for any $y \in \mathbb{R}$ $$ ax - L(x) \leq ax - (xy - f(y)) \ .$$ So in particular for $y = a$ we have $$ ax - L(x) \leq f(a) \ . $$ Since $x \in \mathbb{R}$ was arbitrary we have $$\sup_{x \in \mathbb{R}} (ax - L(x)) \leq f(a) \ .$$ Now I would like to show the reverse inequality. To do this I would like to show that for each $a \in \mathbb{R}$ there exists $x \in \mathbb{R}$ such that $$ f(a) \leq ax - L(x) \ .$$ From this the reverse inequality follows. However I have no idea where to even begin here. I have tried various imitations of the case where $f$ is differentiable, but all of those methods rely heavily on that fact. Also I have noted that I have not been able to really utilize the convexity of $f$ in any of my attempts. In the differentiable case the convexity of $f$ can be used to argue that the supremum is actually a maximum, and that it is achieved at a unique point.
Any ideas? I would prefer a hint or strategy to an outright answer.
One strategy that I thought of is the following. Let $L$ be the Legendre transform of $f$. Now \begin{align} \psi(a) &:= \sup_{b \in \mathbb{R}}(ab - L(b)) = \sup_{b \in \mathbb{R}}(ab - \sup_{c \in \mathbb{R}}(bc - f(c))) \\ &= \sup_{b \in \mathbb{R}}(ab + \inf_{c \in \mathbb{R}} (f(c) - bc)) \\ &= \sup_{b \in \mathbb{R}} \inf_{c \in \mathbb{R}}(f(c) + b(a-c))\,. \end{align} Observe that \begin{align} \inf_{c \in \mathbb{R}} \sup_{b \in \mathbb{R}} (f(c) + b(a-c)) = f(a)\,. \end{align} Now the proof reduces to showing that one can interchange the order of $\sup$ aind $\inf$. This is where the convexity of $f$ comes to use I guess, see for example Sion's minmax theorem. Although this theorem has stronger assumptions, there might exists some analogous theorem that works in this case.