TITLE: Does the derivative of the Legendre transform satisfy $f^{*\prime}(x)=(f')^{-1}(x)$?
QUESTION [1 upvotes]: 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

REPLY [1 votes]: 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}
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