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:

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


 * $\dfrac {\partial^2 g} {\partial p^2} = - \dfrac {\partial^2 f^*} {\partial p^2}$

By Legendre Transform of Strictly Convex Real Function is Strictly Convex and Real Function is Strictly Convex iff Derivative is Strictly Increasing, it holds that:

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 is the Legendre Transform of $f^*$.

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

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