# 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

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

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

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

\(\displaystyle \) | \(=\) | \(\displaystyle x\) | $\quad$ | $\quad$ |

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

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

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

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

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