# $\Omega\subset\subset\mathbb{R}^N$ strongly "analytically convex" at a point $p\in\partial\Omega$ implies that all curvatures at $p$ are positive

I'm reading Convex Analysis by Krantz, and I'm stuck on a remark he makes about "analytic convexity" (which is apparently equivalent to convexity), which he defines as follows:

Let $\Omega\subseteq\mathbb{R}^n$ be a bounded domain with $C^2$ boundary and $\rho$ a defining function for $\Omega$. Fix a point $p\in\partial\Omega$. We say that $\partial\Omega$ is analytically (weakly) convex at $p$ if $$\sum_{j,k=1}^{N}\frac{\partial^2\rho}{\partial x_j\partial x_k}(p)w_jw_k\geq 0,\;\;\;\;\;\;\;\;\;\;\forall w\in T_p(\partial\Omega).$$ We say that $\partial\Omega$ is strongly convex at $p$ if the inequality is strict whenever $w\neq 0$.

The reader may wish to verify that at a strongly convex boundary point, all curvatures are positive (in fact, one may, by the positive definiteness of the matrix $(\partial^2\rho/\partial x_j\partial x_k)$, impose a change of coordinates at $p$ so that the boundary of $\Omega$ agrees with a ball up to and including second order a $p$).

It turns out that I am a reader who wishes to verify this fact. (I'm less interested in the remark in parentheses, but I've included it for the sake of completion.) However I've failed to do this. I've been attempting (and failing) to prove the following case of it all day:

Let $\Omega\subset\subset\mathbb{R}^3$ be an open region with $C^2$ boundary. If $p\in\partial\Omega$ is a strongly convex boundary point (with $\Omega$ having $C^2$ boundary), then the Gaussian curvature of $\partial\Omega$ at $p$, $K(p)$, is positive.

Attempt

Well $\partial\Omega=\rho^{-1}(0)$, so we can invoke the implicit function theorem to get a local parametrization $(u,v)\mapsto (u,v,h(u,v))$, where $h:\mathbb{R}^2\to\mathbb{R}$ is such that $\rho(u,v,h(u,v))=0$. Now (omitting the calculations), $$K(u,v)=\frac{h_{uu}h_{vv}-h_{uv}^2}{(1+h_u^2+h_y^2)^2},$$ and I can relate the derivatives of $\rho$ to the derivatives of $h$ via differentiating both sides of $\rho(u,v,h(u,v))=0$, but after a mess of computations I've failed to get any form which allows me to invoke the definition.

Is there a more simple way to approach this? Am I missing something obvious? I know this can be done using the more geometric definition of convexity, but I'd prefer to use this definition if at all possible. Help of any form is appreciated. Thanks in advance.

• Hint: Think about the symmetric matrix with the entries $h_{uu}, h_{uv}, h_{vv}$. Is it positive definite in view of analytic convexity? What can you conclude about its determinant? Aug 16, 2016 at 6:02
• Well if I show the matrix $$H=\begin{pmatrix}h_{uu} & h_{uv} \\ h_{uv} & h_{vv}\end{pmatrix}$$ to be positive-definite, it would have positive determinant. So $$K(u,v)=\frac{\det H}{(1+h_u^2+h^2_y)^2}>0.$$ But I'm not sure how to show that. I've tried writing the Hessian of $\rho$ (a positive definite matrix) in terms of $h$ (as much as possible that is) and looking at its leading principal minors, but I couldn't see a way through the mess. Am I being dense here? Aug 17, 2016 at 0:07
• Blake: This is just matter of thinking about linear algebra and vector calculus. Can you define positive definite matrix in terms of a quadratic form? Now, think about the 2nd paragraph of your question. Of course, you also need to relate 2nd order derivatives of $\rho$ and of $h$. For this, take $\rho(u,v,z)= z - u-v$, $z=h(u,v)$. Aug 17, 2016 at 0:17
• @studiosus I've posted (what I think is) a proof. Thank you for being patient with me. Aug 17, 2016 at 2:07

I think got it. Thanks @studiosus for being patient with me. Given an arbitrary defining form $\rho:\mathbb{R}^3\to\mathbb{R}$, we can use the implicit function theorem to find a function $h:\mathbb{R}^2\to\mathbb{R}$ such that $\rho\big(x,y,h(x,y)\big)=0$. It follows (as above) that $$K(x,y)=\frac{h_{xx}h_{yy}-h_{xy}^2}{(1+h_x^2+h_y^2)^2}.$$ Now either $(x,y,z)\mapsto h(x,y)-z$ or $(x,y,z)\mapsto z-h(x,y)$ is a (local) defining form for $\Omega$.
Suppose it is the former. Call it $\tilde\rho$. Since the definition of analytic convexity is invariant of choice of defining form, the hessian of $\tilde\rho$ is positive-definite. Since $$\begin{pmatrix}\tilde\rho_{xx} & \tilde\rho_{xy}\\\tilde\rho_{xy} & \tilde\rho_{yy}\end{pmatrix}=\begin{pmatrix}h_{xx} & h_{xy} \\ h_{xy} & h_{yy}\end{pmatrix}$$ is the $2$nd leading principal minor of the hessian of $\tilde\rho$, it has positive determinant; that is, $h_{xx}h_{yy}-h_{xy}^2$ is positive. This is the numerator of the expression for $K(x,y)$. Since the denominator for the expression is always positive, we conclude that $K(x,y)>0$.
If it is the latter, the same procedure applies (you get a negative infront of every entry of the $2$nd leading block matrix but this doesn't affect its determinant).
Thus $K(x,y)>0$.