# How does this prove that Hessian is positive definite?

$\newcommand\pd{\partial}$ I have read through the beginning of this pdf on Légendre transforms. My problem is when it states:

$$\nabla^2g(p)=\nabla x(p)\cdot\nabla^2f(p)\cdot\nabla x(p)>0.$$

The equality convinces me, I just note the nabla denotes a Jacobian (the "gradient" of a vector function) and nabla square is the Hessian. But how does the equality prove that thing is positive definite? Does it hold in general that if $A,B$ are multipliable square matrices and $B$ is symmetric positive definite in our case $A=\nabla x,B=\nabla^2f$), then $ABA$ is p.d.? How would I prove it?

Update:

I read the answer and the problem seems to be a missing transpose on the left gradient. I tried convincing myself of that, but didn't manage. First of all, forget nablas: let me use $H$ for Hessians and $J$ for those "gradients" which are in fact Jacobians. We know $g(p)=g(\nabla f(x(p))$ and $g(p)=J_x(p)=J_x(\nabla f(x(p))$. Call the coordinates $y_j$ to avoid confusion with $x(p)$. Let me use the chain rule to compute second derivatives:

\begin{align*} \frac{\partial^2g}{\partial y_i\partial y_j}(p)={}&\frac{\pd x_j}{\pd y_i}(p)=\sum_{k,\ell=1}^n\frac{\partial x_i}{\partial y_k}(\nabla f(x(p)))\cdot\frac{\pd(\nabla f)_k}{\pd y_\ell}(x(p))\cdot\frac{\pd x_\ell}{\pd y_i}(p)={} \\ {}={}&\sum_{k,\ell=1}^n(J_x(p))_{jk}H_f(x(p))_{k\ell}J_x(p)_{\ell i}=(J_x(p)\cdot H_f(x(p))\cdot J_x(p))_{ji}. \end{align*}

That seems like just the chain rule, and I can't really see where the transpose would appear in those calculations. What am I doing wrong?

• i think the crucial assumption is, that the $f$ is convex and therefore $H(f)\overbrace{=}^{!} \nabla^2 f>0$. then the right hand side constitutes a positive definite bilinear form and everything is fine. – tired Sep 30 '15 at 12:25
• @tired I supposed so, but how does the middle Hessian being p.d. imply the product is too? Look at the second question at the end... – MickG Sep 30 '15 at 12:27
• There are many notational inconsistency in that pdf. You should probably find something easier to read. – user251257 Sep 30 '15 at 12:51
• Notice that $x(p)$ is the unique solution of $p=\nabla f(x)$. By differentiating the equation, you obtain $id = \nabla^2 f(x(p)) \cdot \nabla x(p)$. Thus the Jacobian of $x$ is the inverse of the Hessian of $f$ which is by assumption positive definite and thus invertible. – user251257 Sep 30 '15 at 16:19
• $\nabla x$ is symmetric. So transpose wouldn't change anything. – user251257 Sep 30 '15 at 17:02

We know $\nabla x$ is symmetric since it equals the Hessian of $g$, so:
$$\nabla x(p)\cdot\nabla^2f(p)\cdot\nabla x(p)=\nabla x(p)^T\cdot\nabla^2f(p)\cdot\nabla x(p),$$
but for any matrix $B$ if $A$ is p.d. then $B^TAB$ also is, hence we conclude. Or rather, we reduce the problem to showing $g\in\mathcal{C}^2$, which is done here.