# Convexity of Function implies Convexity of its Legendre Transform

## 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

\(\displaystyle \frac{\d f^*}{\d p}\) | \(=\) | \(\displaystyle -\frac{\d \map f {\map x p} } {\d p} + \frac {\map \d {p \map x p} } {\d p}\) | Definition of Legendre Transform | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -f' \frac {\d x} {\d p} + x + p \frac {\d x} {\d p}\) | Product Rule for Derivatives | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -p \frac {\d x} {\d p} + x + p \frac {\d x} {\d p}\) | Definition of $p$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x\) |

\(\displaystyle \frac {\d^2 f^*} {\d p^2}\) | \(=\) | \(\displaystyle \map {x'} p\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 {\map {p'} x}\) | Derivative of Inverse Function | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 {\map {f''} x}\) | Definition of $p$ | ||||||||||

\(\displaystyle \) | \(>\) | \(\displaystyle 0\) | $\map f x$ is real strictly convex, thus $\map {f'} x$ is strictly increasing, which implies $\map {f''} x > 0$ |

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