Let $u: \mathbb R \to \mathbb R$ be a given function. Then we can consider its upper 1-Lip envelope $$ \hat u(x) \doteq \inf\{g(x) \, \mid\, g \, \text{has Lipschitz constant 1 and}\, g(y) \geq u(y) \, \forall y \in \mathbb R\}. $$ Clearly, since the infimum of two 1-Lip functions, is still a 1-Lip function, the minimization problem above admits a unique minimum in the class of 1-Lipschitz functions (possibly identically $+\infty$). Moreover, it is well know that defining $$ u^{|\cdot|}(x) = \inf_y |x-y| -u(y) $$ one can generate the envelope $\hat u$ via the formula \begin{equation} \hat u(x) = \inf_y |x-y| -u^{|\cdot|}(y) \end{equation}
Is there an analogous second order concept? To make the question more precise, we define an order relation between functions by $g \preceq g^\prime$ if and only $g(y) \leq g^\prime(y)$ for every $y \in \mathbb R$, and we consider the ordered set of functions $$ X \, \doteq \, \{g:\mathbb R \to \mathbb R \, \mid \,|g^{\prime \prime}| \leq 1\, \text{and} \; u \preceq g\}. $$ Since the infimum of two functions that satisfy $|g^{\prime\prime}| \leq 1$ does NOT satisfy the same bound (unlike the first order case), there is not a unique minimal element in $X$ for the relation $\preceq$.
My question is: is there an analogous formula to to the first order one (for $\hat u$) that generates minimal elements in $X$?
