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

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!

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