Legendre Transform of Strictly Convex Real Function is Strictly Convex
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Theorem
Let $\map f x$ be a strictly convex real function.
Then the function $\map {f^*} p$ acquired through the Legendre Transform is also strictly convex.
Proof
| \(\ds \frac {\d f^*} {\d p}\) | \(=\) | \(\ds -\frac {\d \map f {\map x p} } {\d p} + \frac {\map \d {p \map x p} } {\d p}\) | Definition of Legendre Transform | |||||||||||
| \(\ds \) | \(=\) | \(\ds -f' \frac {\d x} {\d p} + x + p \frac {\d x} {\d p}\) | Product Rule for Derivatives | |||||||||||
| \(\ds \) | \(=\) | \(\ds -p \frac {\d x} {\d p} + x + p \frac {\d x} {\d p}\) | Definition of $p$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds x\) |
| \(\ds \frac {\d^2 f^*} {\d p^2}\) | \(=\) | \(\ds \map {x'} p\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {\map {p'} x}\) | Derivative of Inverse Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {\map {f} x}\) | Definition of $p$ |
We have that $\map f x$ is real and strictly convex.
Hence, by Real Function is Strictly Convex iff Derivative is Strictly Increasing, $\map {f'} x$ is strictly increasing.
Then:
- $\map {f} x > 0$
Therefore, the first derivative of $f^*$ is strictly increasing.
By Real Function is Strictly Convex iff Derivative is Strictly Increasing, $f^*$ is strictly convex.
$\blacksquare$
Sources
- 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 4.18$: The Legendre Tranformation