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Let $f\in C^\infty(\mathbb R^n;\mathbb R)$ and assume that

$f(0)=0$, $f(x)>0$ for every $x\in K\setminus \{0\}$ (K compact) and $\partial^2_{i,j}f(0)>0$ ( but $f$ need not be convex in $K$).

I want to show that there exists a quadratic function $q(x)>0$ such that $f(x)\geq q(x)$ for every $x\in K$. Taylor expansion gives

$f(x) = \sum_{i,j} \ f_{i,j}(x) \ x_i x_j$


$f_{i,j}(x) := \int_0^1 (1-s) \ \partial^2_{i,j}f (sx) ds$

so in the strictly convex case one can conclude immediately. What is the fastest way to proceed in the general case? I thought to split $K$ into two parts, $K_1$ containing $0$ where $f$ is strictly convex and $K_2$ (not containing $0$) where $f(x)> C>0$. This gives me quadratic functions $q_1$ and $q_2$ and I take the smallest one.

Thanks for help.

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up vote 1 down vote accepted

I would introduce $g(x)=\dfrac{f(x)}{|x|^2}$ and observe that $\lim\limits_{x\to0}\, g(x)$ exists and is strictly positive, by the positivity of Hessian at $0$. Therefore, $g$ extends to a positive continuous function on $K$. We have $f(x)\ge |x|^2\inf_Kg$ for all $x\in K$.

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