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:
- $\ds \map f x = \max_p \paren {-\map {f^*} p + x p}$
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:
- $\dfrac {\partial^2 g} {\partial p^2} = - \dfrac {\partial^2 f^*} {\partial p^2}$
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.
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 $f$.
$\blacksquare$
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Sources
- 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 4.18$: The Legendre Tranformation