Legendre transformation

Remarks

For $p \in J$ one has $f^*(p) = \sup_{x \in I} \big(p x - f(x)\big)$, and the supremum is attained at $x = g(p)$. Indeed $x \mapsto p x - f(x)$ has derivative $p - f'(x)$, which is positive for $x < g(p)$ and negative for $x > g(p)$. The function $f^*$ is differentiable with $(f^*)' = g = (f')^{-1}$, and $(f^*)^* = f$. The supremum formula, taken over all $x$, defines the convex conjugate of any function, which extends this definition.

Example: $f(x) = e^x$ on $\R$. Here $f'(x) = e^x$ is strictly increasing, $J = (0, \infty)$ and $g(p) = \ln p$. So $f^*(p) = p \ln p - p$ for $p > 0$.

Non-example: $f(x) = x^3$ on $\R$. Its derivative $3x^2$ is not strictly increasing, since $3 \cdot (-1)^2 = 3 \cdot 1^2$, so the definition does not apply.