Convexity of Function implies Convexity of its Legendre Transform

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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

\(\displaystyle \frac{\d f^*}{\d p}\) \(=\) \(\displaystyle -\frac{\d \map f {\map x p} }{ \mathrm d p }+\frac{\d \paren{p \map x p} }{\d p}\) $\quad$ 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