Legendre Transform of Strictly Convex Real Function is Strictly Convex

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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

\(\ds \frac {\d f^*} {\d p}\) \(=\) \(\ds -\frac {\d \map f {\map x p} } {\d p} + \frac {\map \d {p \map x p} } {\d p}\) Definition of Legendre Transform
\(\ds \) \(=\) \(\ds -f' \frac {\d x} {\d p} + x + p \frac {\d x} {\d p}\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds -p \frac {\d x} {\d p} + x + p \frac {\d x} {\d p}\) Definition of $p$
\(\ds \) \(=\) \(\ds x\)
\(\ds \frac {\d^2 f^*} {\d p^2}\) \(=\) \(\ds \map {x'} p\)
\(\ds \) \(=\) \(\ds \frac 1 {\map {p'} x}\) Derivative of Inverse Function
\(\ds \) \(=\) \(\ds \frac 1 {\map {f''} x}\) Definition of $p$


We have that $\map f x$ is real and strictly convex.

Hence, by Real Function is Strictly Convex iff Derivative is Strictly Increasing, $\map {f'} x$ is strictly increasing.

Then:

$\map {f''} x > 0$

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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