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

\(\displaystyle \frac{\d f^*}{\d p}\) \(=\) \(\displaystyle -\frac{\d \map f {\map x p} } {\d p} + \frac {\map \d {p \map x p} } {\d p}\) 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.

$\blacksquare$


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