Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

with

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

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.

share|improve this question
add comment

1 Answer 1

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 $x\in K$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.