# Conditions for Function to be Maximum of its Legendre Transform Two-variable Equivalent

## Theorem

Let $x, p \in \R$.

Let $\map f x$ be a strictly convex real function.

Let $f^*$ be a Legendre transformed $f$.

Let $\map g {x, p} = - \map {f^*} p + x p$

Then:

$\ds \map f x = \max_p \paren {-\map {f^*} p + x p}$

where $\ds \max_p$ maximises the function with respect to a variable $p$.

## Proof

Function $g$ acquires an extremum along $p$, when its first derivative along $p$ vanishes:

 $\ds \frac {\partial g} {\partial p}$ $=$ $\ds -\frac {\partial f^*} {\partial p} + x$ $\ds$ $=$ $\ds 0$ Extremum condition $\ds \leadsto \ \$ $\ds \map { {f^*}'} p$ $=$ $\ds x$

To check if the extremum is a global maximum, consider the second derivative:

$\dfrac {\partial^2 g} {\partial p^2} = - \dfrac {\partial^2 f^*} {\partial p^2}$
 $\ds \frac {\partial^2 f^*} {\partial p^2}$ $>$ $\ds 0$ $\ds \leadsto \ \$ $\ds \frac {\partial^2 g} {\partial p^2}$ $<$ $\ds 0$ $\ds \leadsto \ \$ $\ds \valueat {\frac {\partial^2 g} {\partial p^2} } {\frac {\partial g} {\partial p} \mathop = 0}$ $<$ $\ds 0$

Negative second derivative at the extremum implies extremum being a global maximum.

Therefore:

$\ds \max_p \paren{-\map {f^*} p + x p} = \bigvalueat {\paren {-\map {f^*} p + x p} } {x \mathop = \map { {f^*}'} p}$

The right hand side is the Legendre Transform of $f^*$.

However, $f^*$ is a Legendre transformed $f$.

By Legendre Transform is Involution, this equals $f$.

$\blacksquare$