Existence of Hessian of convex conjugate Define convex conjugate of $f, f^*(x):=\sup_{y\in\mathbb{R}^n}\langle x,y\rangle-f(y)$. Then I want to prove this statement:
Let $f:\mathbb{R}^n\to\mathbb{R}$ be a convex function. Assume $f$ is twice differentiable at $x_0$ and $\det(D^2f)(x_0)=0$. Show that $D^2f^*(Df(x_0))$ doesn't exist.

I have tried to prove it by contradiction:
Since $\det(D^2f)(x_0)=0$, $\exists v\not= 0$, such that $v^TD^2f(x_0)=0$
$f$ is twice differentiable at $x_0$ and $f^*$ is twice differentiable at $Df(x_0)$ implies given $\epsilon >0$, $\exists \delta>0$, such that for all $0<\|x-x_0\| <\delta,0<\|z-Df(x_0)\| <\delta$
$\frac{\|f(x)-f(x_0)-(x-x_0)^TDf(x_0)-\frac{1}{2}(x-x_0)^TD^2f(x_0)(x-x_0)\|}{\|x-x_0\|}<\epsilon\tag{1}$
and,
$\frac{\|f^*(z)-f^*(Df(x_0))-(z-Df(x_0))^TDf^*(Df(x_0))-\frac{1}{2}(z-Df(x_0))^TD^2f^*(Df(x_0))(z-Df(x_0))\|}{\|z-Df(x_0)\|}<\epsilon\tag{2}$
Assume $\|v\|<\delta$, from $(1)$, we get 
$\frac{\|f(x_0+v)-f(x_0)-v^TDf(x_0)\|}{\|v\|}<\epsilon$
Then I don't know what to do. 
I have only find references about the cases where $Df^2(x_0)$ is nonsingular. Could you give me some hints so that I can proceed further or provide me with some references? Thanks! 
 A: We begin with these basic properties of conjugates, which do not require differentiability of $f$ or $f^*$. You can pull these from any basic convex analysis text:
$$f^*(z) = \sup_y \langle z, y \rangle - f(y)$$
$$\partial f^*(z) = \mathop{\textrm{argmax}}_y \langle z, y \rangle - f(y) = \{y~|~z\in\partial f(y)\}.$$
This second condition implies the following:
$$z \in \partial f(y) \quad\Longleftrightarrow\quad y \in \partial f^*(z).$$
The notation $\partial f$ refers to the subdifferential of $f$; this is an
important concept in convex analysis, so do check out a good convex analysis
text (like Rockefellar) to review all of this.
Now to our specific problem. For $y=x_0$, $\partial f(x_0)=\{z_0\}$, where $z_0\triangleq Df(x_0)$. 
Therefore, $x_0\in\partial f^*(z_0)$. Now, if we could confirm that $x_0$ is
the only subgradient—that is, that $\partial f^*(z_0)=\{x_0\}$—then 
we could say that $f^*$ is differentiable at $z_0$, and $Df^*(z_0)=x_0$. But
we cannot! After all, we know that $D^2 f(x_0)$ is not positive definite, so
we cannot be sure that the supremum $\sup_y \langle z_0, y \rangle - f(y)$ has 
a unique solution. 
So it is  possible that $f^*$ is not differentiable at $z_0$. But if that's the
case, then it's not twice differentiable, either, which is exactly what
we seek to prove. So let us assume that $f^*$ is twice differentiable at 
$z_0$; we will show that  still leads to a contradiction.
If $f^*$ is differentiable at $z_0$, then it must be true that $Df^*(z_0)=x_0$.
Substituting back in for $z_0$ gives us this expression:
$$Df^*(Df(x_0))=x_0.$$ 
Differentiating with respect to $x_0$ yields
$$D^2f(x_0) \cdot D^2f^*(Df(x_0)) = I$$
But this is impossible: since $D^2f(x_0)$ is singular, the product 
must be as well; and it cannot possibly be equal to $I$. 
Thus our assumption that $f^*$ is twice differentiable at $z_0$ is contradicted.
Therefore, even though we can't say for sure that $f^*$ is differentiable 
once, we can say for sure that it is not differentiable twice.
