Let $u\in C^\infty(\mathbb{R}^n\times\mathbb{R}^n)$ be symmetric and of strictly positive type on some hypersurface $S \subset \mathbb{R}^n$ diffeomorphic to $\{0\}\times\mathbb{R}^{n-1}$. This means there is some $c>0$, s.t.
$$\forall N>0, \enspace x\_1,...,x\_N \in S,\enspace v\in\mathbb{R}^N: \quad \sum\_{i=1}^N \sum\_{j=1}^N u(x\_i,x\_j)v\_iv\_j \geq c\left(\sum\_{i=1}^N v\_i\right)^2. \quad (\*)$$
Of course this is satisfied for $u$ being uniformly bounded from below $u\geq c>0$. I would guess ($\*$) is merely necessary for boundedness from below but only sufficient on the diagonal ($u(x,x)\geq c$ f.a. $x\in S$) and not on whole $S\times S$. Now I have two questions concerning this kind of functions:


1. Is there some symmetric and smooth $u$ fulfilling ($\*$), that is not bounded from below?
2. Assuming ($\*$) is it possible to show positivity on some open neighborhood $O$ of $S$, i.e.
$$\forall N>0, \enspace z\_1,...,z\_N \in O,\enspace v\in\mathbb{R}^N: \quad \sum\_{i=1}^N \sum\_{j=1}^N u(z\_i,z\_j)v\_iv\_j \geq 0$$
or are there maybe counterexamples?


Thanks a lot!



Strict positive type function on hypersurface also of positive type in neighborhood?
($\*$) yields positivity of the matrices $(u(x\_i,x\_j)-c)\_{i,j=1,...,N}$ for all $N$ and $x\_1,...x\_N\in S$, from where it is not hard to show the following Cauchy-Schwarz-like inequality
$$|u(x,y)-c| \leq \sqrt{u(x,x)-c}\cdot\sqrt{u(y,y)-c}, \qquad x,y\in S.$$
Hence the diagonal somehow controls the off-diagonal and it seems to me, that $u$ can get negative as long as it is sufficiently positive on the diagonal. However I haven't found a counterexample so far.


The answer to the second question is negative. Consider $u(x,y)=-x^1\cdot y^1 + c$ with $x^1,y^1$ being the first components of $x,y\in\mathbb{R}^n$. Then we have $u\equiv c$ on $S\times S$ but the right hand side of $(\*)$ vanishes for $N=2$ and $v=(-1,1)$. Moreover for any neighborhood $O$ of $S$ we find $\varepsilon>\delta>0$ and $\sigma,\rho\in S$, s.t. $(\varepsilon,\sigma),(\delta,\rho)\in O$, but
$$u\big((\varepsilon,\sigma),(\varepsilon,\sigma)\big) + u\big((\delta,\rho),(\delta,\rho)\big) + 2u\big((\varepsilon,\sigma)(\delta,\rho)\big) = -\varepsilon^2-\delta^2+2\varepsilon\delta<0.$$


