0

Let $f$, $g$ be continuous real valued functions defined on some compact space $X$, $g\ge 0$. Show that if $f>0$ on the subset $\{g=0\}$ then there exists a constant $\lambda$ such that $f+ \lambda g> 0$ on $X$ ( the converse is immediate).

Note: this is a standard result. I think I've found it in a book by - Mathematical Economics.

It can be applied to the problem: let $f$ be a quadratic form. In what case is $f$ positive on a subspace $W\subset \mathbb{R}^d$? We consider the $f$ on the sphere $X=S^{d-1}$. Let $g$ be a positive semi-definite quadratic form with zero set $W$.

orangeskid
  • 56,630
  • Aren't quadratic forms continnous real valued fucntions? Is your new $g$ non-negative? – Feng Jun 17 '22 at 03:18
  • What's the definition of 'qudratic form'? In linear algebra, a quadratic form is a real-valued function $f(x)=x^TAx$ for some symmetric matrix $A$. In this case, we can directly apply the standrad result you stated. – Feng Jun 17 '22 at 05:16

1 Answers1

0

$g= 0 \Rightarrow f>0$ implies $f \le 0 \Rightarrow g>0$, so $g$ is strictly positive on the negative subset of $f$ (the problem part). So, there exists an $\epsilon>0$ such that $g\ge \epsilon$ on $\{f\le 0\}$. Also, there exists $M>0$ such that $f>-M$ on $\{f\le 0\}$. We get $$f + \frac{M}{\epsilon} g > -M + \frac{M}{\epsilon} \cdot \epsilon= 0$$

Application: Let $f$ be a quadratic form on $\mathbb{R}^n$ with matrix $A$. Then $f$ is positive definite on the subspace $\sum x_i=0$ if and only if there exists $\lambda>0$ such that $A + \lambda J \succ 0$ ( positive definite). Here $J$ is the $n\times n$ matrix with entries all $1$. The statement can be replaced with : $A + \lambda J$ positive definite for $\lambda >>0$. Now this condition can be stated in terms of (bordered) determinants.

orangeskid
  • 56,630