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

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Let $x, p \in \R$.

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

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

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


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

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


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:

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

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:

\(\ds \frac {\partial^2 f^*} {\partial p^2}\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \frac {\partial^2 g} {\partial p^2}\) \(<\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \valueat {\frac {\partial^2 g} {\partial p^2} } {\frac {\partial g} {\partial p} \mathop = 0}\) \(<\) \(\ds 0\)

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


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

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

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

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