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

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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:

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

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:

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

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 Convexity of Function implies Convexity of its Legendre Transform and Real Function is Strictly Convex iff Derivative is Strictly Increasing

it holds that

\(\displaystyle \frac {\partial^2 f^*} {\partial p^2}\) \(>\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\partial^2 g} {\partial p^2}\) \(<\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\partial^2 g} {\partial p^2}\bigg\rvert_{\frac {\partial g} {\partial p}=0}\) \(<\) \(\displaystyle 0\) $\quad$ $\quad$

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

Therefore:

$\displaystyle \max_p \paren{-\map {f^*} p+x p}=\paren{-\map {f^*} p+x p}\big\rvert_{x=\map {{f^*}'} p}$

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

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

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

$\blacksquare$



Sources