# 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.

$\blacksquare$