Does the derivative of the Legendre transform satisfy $f^{*\prime}(x)=(f')^{-1}(x)$?

Question

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a smooth convex function with an invertible first derivative, define the Legendre transform by \begin{eqnarray} f^*(x):=\max\limits_{p \in [a,b]}\{px-f(p)\} \end{eqnarray} Details can be found in https://en.wikipedia.org/wiki/Legendre_transformation

Then, do we have ${f^*}'(x)=(f')^{-1}(x)?$

if $a=-\infty$ and $b=\infty$ and $f$ has superlinear growth i.e. $\lim\limits_{|x|\rightarrow \infty}\frac{f(x)}{|x|}=\infty$ then we can prove it easily using the fact that, If $f^*(x)=xp-f(p))$ then $p=(f')^{-1}(x).$ (This proof can be found in the above link)

However when $-\infty <a<b< \infty$ we do not have that, because $f^*(x)=px-f(p)$ implies either $p=(f')^{-1}(x)$ or $p=a$ with $x <{f'}(a)$ or $p=b$ with $x >{f'}(b)$ So, I am unable to proceed...Please help

Answer 1

The Legendre transform which is also called conjugate function is not differential, thus we only can obtain the subdifferential: \begin{equation} \partial f\left(x_{0}\right)=\left\{y \in Y: f(x)-f\left(x_{0}\right) \geq\left\langle y, x-x_{0}\right\rangle \quad \text {for all}~ x \in X\right\} \end{equation} In this sense, we have \begin{equation} y \in \partial f(x) \Leftrightarrow x \in \partial f^*(y) \end{equation}