Show that if $f(x,y)\geq a(x^2+y^2)$ for some $a > 0$, $f$ is smooth and $f(0,0) = 0$, then Gaussian curvature at $(0,0)$ positive There is a question in Differential Geometry which I am not sure how to solve:
For $f:\mathbb{R}^{2}\to\mathbb{R}$, show that if $f(x,y) \geq a(x^2+y^2)$ for some constant $a > 0$, $f$ is a smooth function,
and $f(0,0) = 0$, then Gaussian curvature at $(0,0)$ is positive.
I have tried to explain geometrically that the surface of $f$
looks like a sphere at $(0,0)$ and its a minimum point,
because the Gaussian curvature is the determinant of the Hessian function at critical point, therefore it need to be positive.
But I am not sure if this solution is formal enough (because it might be zero as well).
I also tried to solve it with Taylor series of $2$ variables.
I will really appreciate it if you can help.
 A: First: the condition $f(x,y) \ge a(x^2 + y^2)$ is enough to be satisfied only in a region $\{ x^2 + y^2 < \delta^2\}$.
Now, from the above we conclude that $f$ has a local minimum at $(0,0)$. If the function $f$ has partial derivatives at $(0,0)$, they must be $0$.
Now, assume that $f$ is $C^2$ around $(0,0)$ ( it is enough that $f$ has the second differential at $0$. Let $H$ be the Hessian matrix at $(0,0)$. Then for every $v \in \mathbb{R}^2$ ( a unit vector) the function $\phi(t) \colon t \mapsto f(t\cdot v)$ is smooth around $0$, its  derivative at $0$ is $0$, and its second derivative at $0$ is
$$\phi^{(2)}(0) = v^t \cdot H_f(0,0) \cdot v = \sum_{i,j} \frac{\partial^2 f}{\partial x_i \partial x_j} v_i v_j$$
Now, we have the approximation formula valid for any $C^2$ function on $\mathbb{R}$
$$\phi(t) = \phi(0) + \phi'(0)t + \frac{1}{2}\phi^{(2)}(0)\cdot  t^2 + \epsilon(t) \cdot t^2$$
with $\epsilon(t) \to 0$ as $t\to 0$.
We get
$$f(t v) = \frac{1}{2} \sum_{i,j} \frac{\partial^2 f}{\partial x_i \partial x_j} v_i v_j \cdot t^2+ \epsilon(t) \cdot t^2$$
However, we have $f(t v) \ge a\cdot \|t v\|^2 = a t^2$ ( we took $v$ a unit vector). We conclude
$$\frac{1}{2} \sum_{i,j} \frac{\partial^2 f}{\partial x_i \partial x_j} v_i v_j \cdot t^2+ \epsilon(t) \cdot t^2\ge a t^2$$
that is
$$\frac{1}{2} \sum_{i,j} \frac{\partial^2 f}{\partial x_i \partial x_j} v_i v_j\ge a - \epsilon(t)$$
and since $t$ is arbitrary, and we could take $t\to 0$, with the effect of $\epsilon(t) \to 0$, we get
$$ \sum_{i,j} \frac{\partial^2 f}{\partial x_i \partial x_j} v_i v_j\ge 2 a$$
for all unit vectors $v= (v_i)$.  This means that the Hessian is positive definite at $0$. Now, the curvature of the graph of $f$ at a point is given by
$$K = \frac{F_{xx} F_{yy} - F_{xy}^2}{ (1 + F_x^2 + F_y^2)^2}$$
and the numerator at $(0,0)$, being the  determinant (of the) Hessian, it's positive from the above ( a positive definite matrix has positive determinant).
Note:

*

*the reasoning above works for any $a$, before the conclusion "Hessian positive definite". If $a=0$ ( that is, $(0,0)$ is a local minimum), then we get $H_{f, (0,0)}$ positive semi-definite, so its determinant is $\ge 0$. This implies the curvature is $\ge 0$.


*Perhaps it's better to write the condition $f(x,y) \ge \frac{1}{2} k ( x^2+ y^2)$. Then the Hessian at $(0,0)$ is $\succeq k \cdot I_2$ ( in the positive semi-definite order), and so its determinant is $\ge k^2$. Moreover, the partials at $0$ are $0$. We get $K_{(0,0)} \ge k^2$.


*Intuitively: curvature positive at a point of a graph if and only if the graph is on one side of the tangent plane ) close to that point), curvature negative if and only if it's on "both sides".
