Strict positive type function on hypersurface also of positive type in neighborhood?

Question

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!

Answer 1

($*$) 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 $$u\big((\varepsilon,\sigma),(\varepsilon,\sigma)\big) + u\big((\delta,\rho),(\delta,\rho)\big) + 2u\big((\varepsilon,\sigma)(\delta,\rho)\big) = -\varepsilon^2-\delta^2+2\varepsilon\delta<0.$$