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

Theorem
Let ${x, p}\in \R$.

Let $f \left({ x } \right)$ be a strictly convex real function.

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

Let $ g \left({ x, p } \right)= - f^* \left({ p } \right) + x p$

Then $ \displaystyle f \left({ x } \right) = \max_p \left({ - f^* \left({ p } \right) + x p } \right)$, where $ \displaystyle \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:


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

By Convexity of Function implies Convexity of its Legendre Transform 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


 * $ \displaystyle \max_p \left({ - f^* \left({ p } \right) + x p } \right)=\left({ - f^* \left({ p } \right) + x p } \right) \big \rvert_{x= {f^*}' \left({ p } \right)}$

The RHS is the Legendre Transform of $f^*$.

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

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